Parabola calculator Print parabola calculator
( x )2 = ( y )
x2 + x + y + = 0
x2 + xy + y2 + x + y + = 0
Parabola figure - 1
Focus
( , )
Vertex
( , )
Directrix equation
Symmetry axis
Parabola angle
Input limit:
Notes:
2
                  Examples:
Parabola equation:
Parabola and line intersection
x + y + = 0
Intersection: Point 1
Point 2
Tangent line at 1
At point 2
Check if a point is inside of a parabola
Point (x , y): ( , )
Parabola with center at (0,0) Inclined parabola Tangent line ex: 4 , 4a , 4b
Parabola tables Foci vertex ex: 1 , 1a , 1b , 1c , 1d Inclined parabola ex: 5 , 5a , 5b
Parabola tangent lines Parabola equation ex: 2 , 2a , 2b , 2c Point location ex: 6 , 6a , 6b
Parabola and line intersection Parabola and line ex: 3 , 3a , 3b , 3c , 3d
Parabola -   summary Print hyperbola summary
Parabola figure - 1
A parabola is the locus of all points whose distances from a fixed point equal their distance from a fixed line called the directrix, the fixed point is the focus
We set the x axis tangent to the parabola.
According to the parabola definition we have:
|FP| = |PA|       |FQ| = |QB|       |FS| = |SD|       |FT| = |TE|
Parabola figure - 1
We denote the distance from the focus to the origin as
p,  then from the distance formula we have   |FP = PA|
FP=√((0-x)^2+(p/2-y)^2 )
PA=|y-p/2|
After squaring both sides we get:
x^2+(p/2-y)^2=y-p/2
And finally we get the general equation of a parabola: x22py = 0 notice that   p > 0
Notice that this parabola has the vertex at the origin and focus distance of  p  from the directrix.
If the vertex is located at  (h , k) then the parabola equation becomes:
Vertical parabola (x − h)2 = 2p(y − k)
Horizontal parabola: (y − k)2 = 2p(x − h)
The most recognizable form of a parabola is: y = ax2 + bx + c
This equation can be presented by the form: (y − k) = n(x − h)2 (see examples 1 - 1c)
Notes:
1. The values of  h  and  k  presents the coordinate of the vertex.
2. if the value of  h  is negative the vertex is at the positive location and if  k  is negative the  y  axis is at the positive location.
3. if  x  is the square value then the parabola is vertical, open side direction is up or down.
4. if  y  is the square value then the parabola is horizontal open side is to the left or right.
5. The sign of  n  is responsible for the direction of the open side of the parabola.
Positive values are for up and right directions.
Negative values are for down and left directions.
Vertex at (0,0) summary           p > 0 Print hyperbola summary
Type Sketch parabola equation Focus at
Vertex
Directrix
Symmetry
1 Parabola figure - 3 x2 = 2py
(0 , p/2)
(0 , 0)
y = −p/2
x = 0
2 Parabola figure - 4 x2 =2py
(0 , − p/2)
(0 , 0)
y = p/2
x = 0
3 Parabola figure - 1 y2 = 2px
(p/2 , 0)
(0 , 0)
x = −p/2
y = 0
4 Parabola figure - 2 y2 =2px
(−p/2 , 0)
(0 , 0)
x = p/2
y = 0
Vertex at (a,b) summary           p > 0 Print hyperbola summary
Type sketch parabola equation Focus at
Vertex
Directrix
Symmetry
1 Parabola figure - 3 (x − a)2 = 2p(y − b)
(a , b + p/2)
(a , b - p/2)
y = b − p/2
x = a
2 Parabola figure - 4 (x − a)2 =2p(y − b)
(a , b − p/2)
(a , b + p/2)
y = b + p/2
x = a
3 Parabola figure - 1 (y − b)2 = 2p(x − a)
(a + p/2 , b)
(b - p/2 , b)
x = a − p/2
y = b
4 Parabola figure - 2 (y − b)2 = −2p(x − a)
(a − p/2 , b)
(b + p/2 , b)
x = a + p/2
y = b
Parabola − tangent lines Print hyperbola summary
Parabola tangent line
We will take the general equation of a parabola as:
(x − a)2 = 2p(y − b)
The slope of the tangent line can be found by implicit derivation of the equation.
2(x-a)=2p dy/dx
dy/dx=(x-a)/p
and the tangent line equation is:
y=(x_1-a)/p x+y_1-(x_1-a)/p
If the parabola is given by the equation: Ax2 + Dx + Ey + F = 0
The slope by implicit derivation is: dy/dx=m=(-2Ax-D)/E
And the tangent line equation at (x1 , y1) y=(-2Ax_1-D)/E x+y_1+(2Ax+D)/E x_1
Parabola tangent line
The general form of a horizontal parabola is.
(y − b)2 = 2p(x − a)
By implicit derivation we get the slope
m=dy/dx=p/(y-b)
and the tangent line equation is:
y=p/(y_1-b) x+y_1-p/(y_1-b) x_1
If the parabola is given by the equation: Cy2 + Dx + Ey + F = 0
The slope by implicit derivation is: 2Cy dy/dx+D+E dy/dx=0
2Cy dy/dx+D+E dy/dx=0
And the tangent line equation at (x1 , y1) y=(-D)/(2Cy+E) x+y_1+D/(2Cy+E) x_1
Parabola tangent line
The general form of a tilted parabola is.
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
By implicit derivation we get the slope
2A+Bx dy/dx+2Cy dy/dx+C+E dy/dx=0
m=dy/dx=(-2Ax-By-D)/(Bx+2Cy+E)
and the tangent line equation at point  (x1 , y1)  is:
y=(-2Ax-By-D)/(Bx+2Cy+E) x+y_1-(-2Ax-By-D)/(Bx+2Cy+E) x_1
Parabola and line intersection Print parabola and line intersection summary
Parabola and line intersection
If the parabola and the line are given by the equations:
Ax2 + Dx + Ey + F = 0(1)
mx + ny + q = 0(2)
Solving this equations we get the intersection points.
From equation (2) we have
y=(-mx-q)/n(3)
Substitute the value of  y  into eq.  (1)  to get  x.
Anx^2+Dnx-Emx-Eq+nF=0
After multiplying all terms by B we have: Anx2 + (Dn − Em)x − Eq + nF = 0
And  x  is equal to:x_1,2=(Em-Dn±√((Dn-Em)^2+4An(Eq-nF) ))/2An
The value of  y  is found according to equation  (3)
Because the value of the square root should be greater then  0  for two solutions we have
Total intersections Condition
Two points (Dn − Em)2 + 4An(Eq − Fn) > 0
One point (tangent) (Dn − Em)2 + 4An(Eq − Fn) = 0
No intersection (Dn − Em)2 + 4An(Eq − Fn) < 0
Parabola and line intersection If the parabola is horizontal then the equation of the parabola is:
Cy2 + Dx + Ey + F = 0(4)
mx + ny + q = 0(5)
Solving this equations we get the intersection points.
From equation  (5)  we have
x=(-ny-q)/m(6)
Substitute the value of  x  to eq.  (4)  to get  y.
Cy^2+D((-ny-q)/m)+Ey+F=0(7)
After multiplying all terms by  B  we get: Cmy2 + (Em − Dn)y − Dq + Fm = 0
And  y  is equal to: y_1,2=(Dn-Em±√((Em-Dn)^2+4Cm(Dq-Fm) ))/2Cm
The value of  x  is found according to equation  (6)
The conditions for intersections are:
Total intersections Condition
Two points (Em − Dn)2 + 4Cm(Dq − Fm) > 0
One point (tangent) (Em − Dn)2 + 4Cm(Dq − Fm) = 0
No intersection (Em − Dn)2 + 4Cm(Dq − Fm) < 0
Parabola and line intersection If the parabola is inclined then the equation of the parabola is:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0(8)
mx + ny + q = 0(9)
Solving this equations we get the intersection points.
From equation  (9)  we have
x=(-ny-q)/m(10)
Substitute the value of  x  into eq.  (8)  to get  y.
Ax^2+Bx((-mx-q)/n)+C((-mx-q)/n)^2+Dx+E((-mx-q)/n)+F=0 (11)
After solving the value of  x  is: x=(-t±√(t^2-4zr))/2z
where: z = An2 − Bmn + Cm2
t = 2Cmq − Bnq + Dn2 − Emn
r = Cq2 − Enq + n2F
Total intersections Condition
Two points t24zr > 0
One point (tangent) t24zr = 0
No intersection t24zr < 0
Inclined parabola Print inclined parabola summary
Inclined parabola
According to the parabola definition we have
Distances: f p = p a
FP=√((x-f_x )^2+(y-f_y )^2 )
From the equation of the distance from a point to line we have the expression:
PA=|mx+ny+q|/√(m^2+n^2)
√((x-f_x )^2+(y-f_y )^2 )=|mx+ny+q|/√(m^2+n^2 )
After multiplying all the terms we get the general equation of a tilted parabola.
Notice that we have now the square values of  x  and  y  and also the term  xy  added to the parabola (see examples 5, 5a, 5b).
n^2 x^2-2mnxy+m^2 y^2-2(m^2 f_x+p^2 f_x+md)x-2(m^2 f_y+p^2 f_y+nd)y+(m^2+n^2 )(〖f_x〗^2+〖f_y〗^2 )-d^2=0
If the equation of the parabola is: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 then.
A = n2(1)
B = −2mn(2)
C = m2(3)
D = −2(m2fx + n2fx + mq)(4)
E = −2(m2fy + n2fy + nq)(5)
F = (m2 + n2 )(fx2 + fy2) − q2(6)
From equations (1), (2) and (3) B=-2√AC or B24AC = 0
We can see that the signs of  A  and  C  must be positive (equals to square values).
The term B is a function of the slope of the directrix:   mx +ny + q = 0
We can see from eq. (2) that the sign of the term  B  is also the sign of the directrix slope.
m=±√C(7)
n=±√A(8)
q^2 (4C+4A-1)+q(4D√C+4E√A)+D^2+E^2-4F(A+C)=0(9)
f_x=(-2mq-D)/2(m^2+p^2 ) =(-2q√C-D)/2(A+C)(10)
f_y=(-2pq-E)/2(m^2+p^2 ) =(-2q√A-E)/2(A+C)(11)
From the definition of the parabola we can see that if a point fulfills the location conditions:
If fp = pa point is on the parabola
If fp < pa point is inside the parabola
If fp > pa point is outside of the parabola
Example 1 - vertices and foci Print vertices and foci example 1
Given the parabola   x2 + 2y − 3x + 5 = 0,   find the vertex, focus, directrix and the axis of symmetry.
By the method of completing the square we write the equation of the parabola as:
x23x       = −2y − 5
(x-3/2)^2-9/4=-2y-5
(x-3/2)^2=-2(y+11/8)
From the last equation we notice that   a = 3/2    b =11/8   and p = 1
Sketch ex 1
Because the type of parabola is:   x2 = −py   the parabola is horizontal and the symmetry axis is at the  y  direction:
And the vertex is at    (a , b)  =  (3/2 , −11/8)
The focus coordinate is at:
(a , b - p)  =  (3/2 , −11/8 - 1/2) =  (3/2 , −15/8)
The directrix is presented by the line:
y = b + p/2 =11/8 + 1/2 = −7/8
The axis of symmetry is at:    x = 3/2    (see sketch at left).
Example 1a - vertices and foci Print vertices and foci example
Given the parabola   y24y + 2x + 7 = 0,   find the vertex, focus, directrix and axis of symmetry.
We will solve the general solution of the parabola equation:       y2 + Ay + Bx + C = 0
By the method of completing the square we write the equation of the parabola as:
y2 + Ay       = −Bx −C y24y = −2x − 7
(y+A/2)^2-A^2/4=-Bx-C (y − 2)24 = −2x − 7
(y+A/2)^2=-B(x+A^2/4B-C/B) (y − 2)2 = −2(x + 1.5)
From the parabola given values we have:         A =4         B = 2         C = 7
From the last equation we see that the coordinate of the vertex is:
(a ,b)=(A^2/4B-C/B   ,A/2)
The distance from focus to directrix is: p=|-B|/2=2/2=1
The sign of  B  is negative so the parabola open side is to the left direction.
Sketch ex 1
Vertex to directrix distance is     p = 1/2 = 0.5
The focus coordinate is at:
(C/B-A^2/4B-p/2   ,A/2)=(7/2-16/8-1/2,(-4)/2)=(1,-2)
The directrix is passing through the point:
(a + p/2 , b) = (1.5 + 0.5 , 2) = (−1 , 2)
The equation of the directrix is   x =1
The axis of symmetry is at:    y = 2    (see sketch at left).
Example 1b - vertices and foci Print vertices and fovi example
Find the coordinate of the foci and the directrix equation of the parabola given by the equation
y2 =12x.
Sketch ex 1
We mark the distance between the directrix and the focus by  p  then       y2 =12x =2px       and       p=12/2 = 6
From the equation of the parabola   (y − 0)2 = −12(x − 0)     We can see that     a = 0   and   b =0
Because  y  is squared the parabola is in the horizontal direction and because the sign of  x  is negative the open direction of the parabola is to the left (see sketch).
And the focus is at:    (3 , 0)
The vertex is at:    (a , b) = (0 , 0)
The directrix is a verical line passing through the point:
(3 , 0) and the equation of the directrix is:
x = 3
The symmetry axis of the parabola is at    y = 0
Example 1c - vertices and foci Print vertices and foci example
Given the parabola   x2 − 2x − 2y + 3 = 0,   find the vertex, focus, directrix and axis of symmetry.
Sketch ex 1
By the method of completing the square we write the equation of the parabola as:
(x2 2x)       = 2y − 3
(x − 1)21 = 2y − 3
(x − 1)2 = 2(y − 1)
The general parabola equation is:
(x − a) = 2p(y − 1)
We see that   a = −1    b = −1    and    p = 1
Because the value of  2p  is positive, the parabola is vertical:   and the symmetry axis is in the  y  direction:
The vertex is at    (a , b)  =  (1 , 1)
Notice that the sign of the coordinates values are opposite to the values of  a  and  b .
The focus coordinate is at: (a , b + p/2)  =  (1 , 1 + 1/2) =  (1 , 1.5)
The directrix is at the line: y = (b − p/2) = 1 − 0.5 = 0.5
The axis of symmetry is at: x = a = 1    (see sketch above).
Example 1d - foci and directrix Print foci and directrix example
Given the directrix line of a parabola by the equation  y = 1.75  and the focus at the point
F(2 , 0.25).  Find the equation of the parabola and the vertex and the symmetry line equations.
Sketch of example 1d
The y value of the directrix is 1.75 and the y value of the focus is 0.25 then the y component of the vertex is:
v_y=(d_y+f_y)/2=(1.75+0.25)/2=1
The x value of the vertex is the same as the x value of the focus
vx = fx = −2
The coordinate of the vertex is:     v(−2 , 1)
The distance from the vertical coordinate of the directrix to to the vertical coordinate of the focus is:
p = dy − fy = 1.75 − 0.25 = 1.5
Because the y coordinate of the directrix is greater then the y coordinate of the focuse the parabola is vertical the open direction is pointing down.
The general equation of a parabola is: (x − h)2 = 2p(y − k)
The value of h = −2 ,     k = 1 ,     p =2 * 1.5 = − 3 (x + 2)2 = −3(y − 1)
After openning the paranthesis we get the expression: x2 + 4x + 3y + 1 = 0
The symmetry line equation is (opposite sign h = x =2): x =2
Example 1e - tilted foci and directrix Print foci and directrix example
Given the focus of a parabola at  (1 , 4) and the directrix equation  x + y − 9 = 0  find the equation of the parabola and the coordinates of  (xd , yd).
Sketch of example 1d
According to the parabola definition the distanc from the focus to any point on the parabola denoted by  (x , y) is equal to the distance from the point to the directrix line.
√((f_x-x)^2+(f_y-y)^2 )
The distance from x and y to the directrix is:
|mx+ny+q|/√(m^2+n^2 )
From the given values we have:
fx = 1 fy = 4 m = 1 n = 1 q = 9
√((1-x)^2+(4-y)^2 )=|x+y-9|/√(1+1)
After squaring both sides we get:
(1-x)^2+(4-y)^2=(x+y-9)^2/2
And after a few steps of simple algebra we get the final equation of the parabola
x22xy + y2 + 14x + 2y − 47 = 0
Slope of the directrix is: m_d=-m/n slope of symmetry line is: M_d=n/m
The equation of the symmetry line is: Mdx − y + fy − Mdfx = 0
Point (dx , dy) is the intersection of both lines. mx + ny + q = 0
So we have to solve the set of the following linear equations: (■(M_d&-1@m&n)){■(x@y)}={■(M_d f_x-f_y@-q)}
And the value of the intersection of symmetry line and the directrix  (dx , dy)  is:
x_d=|■(M_d f_x-f_y&-1@-q&n)|/|■(M_d&-1@m&n)| =(M_d f_x n-nf_y-q)/(M_d n+m)
y_d=|■(M_d&M_d f_x-f_y@m&-q)|/|■(M_d&-1@m&n)| =(-qM_d+mf_y-M_d f_x m)/(M_d n+m)
The coordinate of the vertex   (vx , vy)   is:
(v_x  ,v_y )=((d_x+f_x)/2   ,(d_y+f_y)/2)=((3+1)/2  ,(6+4)/2)=(2 ,5)
Example 2 - parabola equation Print vertices and foci example
The focus of a parabola is located at the point  (0 , 2)  and the vertex at  (0 , 4)  find the
equation of the parabola.
Sketch ex 2
We see that the focus and the vertex are located along the y axis that mean that the parabola is vertical and because the vertex is above the focus the open side of the parabola is downward.
The directrix coordinate is:   (x , 6)
From the definition of the parabola we have:
d1 = d2
√((x-0)^2+(y-2)^2 )=|y-6|
After squaring both sides we get
(x − 0)2 + (y − 2)2 = (y − 6)2
x2 + y24y + 4 = y212y + 36
After arranging terms we get the equation of the parabola: x2 + 8y − 32 = 0
Example 2a - parabola equation Print vertices and foci example
The vertex of a parabola is located at the point   (−1.5 , 2)   and the directrix equation is
x = −1  find the cordinate of the vertex and the equation of the parabola.
Sketch ex 2a
Sketch ex 2a
The directrix line is at  x =1  (vertical line) this means that the parabola is horizontal.
Because the vertex  x  coordinate is at  
x =1.5  the opening side of the parabola is to the left.
The symmetry axis is at  y = 2   (see sketch)
The directrix and the symmetry axis intersection coordinate is at:   (−1 , 2). In order to get a general solution we will mark this coordinate value as  (d , y).
From the definition of the parabola we have:
d1 = d2
√((F_x-x)^2+(F_y-y)^2 )=|d-x|
After squaring both sides we get
(Fx − x)2 + (Fy − y)2 = (d − x)2
Fx22Fxx + x2 + Fy22Fyy + y2 = d22dx + x2
After arranging terms we get the equation of the parabola:
y22Fyy + (2d − 2Fx)x + Fx2 + Fy2 − d2 = 0
In our case: Fx = −2 Fy = 2 d =1
Substitute values to the equation: y22 ‧ 2y + [2(− 1)2(− 2)]x + 4 + 4 − 1 = 0
And the parabola equation is: y24y + 2x + 7 = 0
Example 2b - parabola equation Print vertices and foci example
Find the transformation equations from the polinomial form   x2 + ax + by + c = 0   to the standard form of the parabola  (x + h)2 = p(y + k)  and vice verse.
From:           x2 + ax + by + c = 0
(x2 + ax) + by + c = 0
Now we will change the paranthasis value into square expression
(x+a/2)^2-a^2/4+by+c=0
(x+a/2)^2=-b(y-a^2/4b-c/b)
From the last expression we can see that
From the last expression we can see that h=a/2 k=-a^2/4b-c/b
And the coordinate of the vertex is (-a/2,a^2/4b+c/b)
p is twice the diatance from the focus to the directrix line         p = |b| / 2
The open side of the parabola is pointing down if  b  is negative and pointing up if  b  is posotive.
From:           (x + h)2 = p(y + k)
After multiplying terms we get:
x2 + 2hx + h2 = py + pk
After arranging terms we get:
x2 + 2hx − py + h2 − pk = 0
the transformation values are: a = 2h b = − p c = h2 − pk
Example 2c - parabola equation Print vertices and foci example
The focus of a parabola is located at the point  (1.5 , 1)  and the vertex at  (1 , 1)  find the
equation of the parabola the equation of the directrix and the equation of the symmetry line.
Sketch ex 2
We notice that the  y  coordinate of both points the focus and the vertex are the same so the open side of the parabola is directed to the left or right. Because the focus  x  point is less then the  x  axis of the vertex the open direction is to the left (see sketch).
The distance between focus and vertex is:
p = 2|fx − vx | = 2|1.5(1)| = 1
and the point dx on the directrix is:
dx = vx + p =1 + 1 =0.5
The equation of the parabola is:
(y − vy)2 =2p(x − vx)
(y − 1)2 = 2(x + 1)
After opening parenthesis we get another form of the parabola: y2 + 2x − 2y + 3 = 0
The directrix line is vertical and is passing through the point (−0.5 , 1)         x = − 0.5
And the symmetry line equation from the sketch is:         y = 1
Example 3 - parabola and line intersection Print vertices and foci example
Find the general expression for the intersection of a parabola   x2 + Ax + By + C = 0
and the line   y = mx + q
With the expression developed find the intersection of the following parabola and line
x2 + 2x − 4y + 3 = 0   and the line   x + 3y − 5.5 = 0.
Substitute  y  from the line equation into the parabola equation to find the value of  x:
x2 + ( A + Bm)x + Bq + C = 0
and  x  coordinates are the solutions of this quadratic equation.
Sketch ex 3
From line equation we have:
y=-1/3 x+5.5/3
Substitute y into parabola equation we get.
x^2+2x-4((-x+5.5)/3)+3=0
3x2 + 6x + 4x − 22 + 9 = 0
3x2 + 10x − 13 = 0
x_1,2=(-10±√256)/6=-4.33,1
The values of y are: y_1=(4.33+5.5)/3=3.28 y_2=(-1+5.5)/3=1.5
And the intersection points are   (−4.33 , 3.28)   and   (1 , 1.5)
Example 3a - parabola and line intersection Print vertices and foci example
Find the intersection points of a parabola   (y − 1)22(x + 2.5) = 0   and the lines   a)  x = 0 and  b)  the line   y = 0.
Sketch ex 3a
Because  y  is the square item and the sign of  x  is negative the opening of the parabola is to the right direction (see summary table)..
From the parabola equation we have:
y22y − 2x − 4 = 0
a) Substitute line equation   x = 0   into the parabola equation we get.
y22y − 4 = 0
The solutions of this equation are:  
y_1,2=1±√5=3.24 ,-1.24
And the intersection points are    (0 , −1.24)    and    (0 , 3.24)
b)   Substitute the line value  y = 0  to the parabola equation we get:   −2x − 4 = 0
It can be seen that there is only one solution for  x  and it is at   x = −2   therefore only one intersection point exists at   (−2 , 0)
Example 3b - parabola and line intersection Print vertices and foci example
Find the intersection points of the a parabola   x22xy + y2 + 10x + 22y 71= 0   and the lines   a)  x + y − 1 = 0.   b)  x = 4.   c)  y = 0. Also derive a general expression for the intersection points.
a) Substitute the value of  y  from the line equation   y = − x  + 1   into the parabola equation.
x22x (− x + 1) + (− x + 1)2 + 10x + 22( − x + 1)71 = 0
Sketch ex 3b
After arranging terms we get the quadratic equation:
x24x − 12 = 0
x=(4±√(16+48))/2=(4±8)/2=6 ,-2
And the values of  y  coordinates are:
y1 = − 6 + 1 = − 1 = − 5
y2 = 2 + 1 = 3
Finally the intersection coordinates are:
(6 ,5) (−2 , 3)
b) Substitute x = 4 into the parabola equation:           16 − 8y + y2 + 40 + 22y −71 = 0
y2 + 14y − 15 = 0
y=(-14±√(196+60))/2=(-14±16)/2=1 ,-15
And the intersection points are: (4 , 1) and (4 , −15)
c) Substitute y = 0 into the parabola equation:           x2 + 10x −71 = 0
x=(-10±√(100+284))/2=(-10±19.6)/2=4.8 ,-14.8
And the intersection points are: (4.8 , 0) and (−14.8 , 0)
The general expression for the intersection points of:
Parabols Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 (1)
Line mx + ny + q = 0 (2)
From equation (2) we get the value of  y: y=-m/n x-q/n (n ≠ 0)
Substitute the value of  y  into the equation of the parabola to get the values of  x:
A(-n/m y-d/m)^2+B(-n/m y-d/m)y+Cy^2+D(-n/m y-d/m)+Ey+F=0
(An^2)/m^2  y^2+2A nd/m^2 +(Ad^2)/m^2 -Bn/m y^2-Bd/m y+Cy^2-Dn/m y-Dd/m+Ey+F=0
((An^2)/m-Bn+Cm) y^2+(Em-Bd-Dn)y+2A nd/m+(Ad^2)/m-Dd+Fm=0
We got a quadratic equation and the solutions are the values of  x.   The values of  y  can be found according to equation (2) .
From equation (2) we get the value of  x: x=-n/m x-q/m (m ≠ 0)
Substitute the value of  x  into the equation of the parabola to get the values of  x:
A(-n/m y-d/m)^2+B(-n/m y-d/m)y+Cy^2+D(-n/m y-d/m)+Ey+F=0
(An^2)/m^2  y^2+2Ad/m^2  y+(Ad^2)/m^2 -Bn/m y^2-Bd/m y+Cy^2-Dn/m y-Dd/m+Ey+F=0
((An^2)/m^2 -Bn/m+C) y^2+(2Ad/m^2 -Bd/m-Dn/m+E)y+(Ad^2)/m^2 -Dd/m+F=0
We got a quadratic equation and the solutions are the values of  y.   The values of  x  can be found according to equation (2) .
Example 3c - parabola and line intersection Print vertices and foci example
Find the intersection points of the parabola   (y − 2)2 = 4(x − 1)   and the lines
2x + y − 8 = 0.   Also find the tangent lines at the intersection points.
Sketch ex 3b
The given parabola can be presented by the equation.
y24y − 4x + 8 = 0
From the line equation we have:
x = 4 − y/2
Substitute  x  into the parabola equation.
y24y − 4(4 − y/2) + 8 = 0
y22y − 8 = 0
The solution for  y  are: y=(9±√(81-40))/2=(9-6.4)/2=7.7  ,1.3
The value of x are: x1 = 44/2 = 2 x2 = 4 + 2/2 = 5
And the intersection coordinates are: (2 , 4) (5 , −2)
The slopes of the tangent lines by explicit differentiation are: dy/dx=m=1/(2y_1,2-6)
dy/dx=m=4/(2y_1,2-4)
m_1=1/(2∙7.7-6)=0.106 m_2=1/(2∙1.3-6)=-0.294
Line equation when a point and the slope are given is:             y = mx + y1 − mx1
The first tangent line: y = x + 4 − 2 x − y + 2 = 0
The second tangent line: y =0.5x + (2) + 0.5 • 5 x + 2y −1 = 0
Example 3d - Line perpendicular to parabola Line perpendicular to parabola example
Find the general equation of the line perpendicular to the parabola given by the equation  
(x + h)2 = p(y + k)   Then use the equation to find the perprndicular line to the parabola
(x − 4)2 = −3(y + 2)  at the point  (1 , −5).
Sketch ex 3d
The perpendicular line at any point on the parabola is also perpendicular to the tangent line.
The tangent line by explicit derivation is:
2(x+h)=p dy/dx
dy/dx=m=2/p (x+h)
Line tangent to the parabola at point (x1 , y1) is:
2(x1 + h)x − py + py12(x1 + h) x1 = 0
2x − y − 7 = 0
Slope of the line perpendicular to the parabola is:
M=-1/m=(-p)/(2(x)+h)
If the tangency point is at   (x1 , y1)   then the equation of the perpendicular line is:
y = Mx + (y1 − Mx1) where M=(-p)/2(x_1+h)
Substitute the value of M we get the expression: p/2(x_1+h)  x+y-[y_1+(px_1)/2(x_1+h) ]=0
After multiplying terms we get: px+2(x_1+h)y-2(x_1+h) y_1-px_1=0 (1)
Now substitute the values: h =4 k = 0 p =3 x1 = 1 y1 =5
We get the perpendicular line to the parabola as: x + 2y + 9 = 0
Notice that the vertex is at point       v(4 , −2)
and focus to directrix distance is p = |−3 / 2| = 1.5
because 2p =3 (negative) the open side of the parabola is down.
The focus is at   f(4 ,20.75) = f( 4 , 2.75)
The directrix line is at:     y =1.25     and the symmetry line is:     x = 4
Example 4 - tangent line Print parabola tangent lines example
A parabola is given by    y22y −2x − 4 = 0,    find,  a) the equation of the tangent line at a point on the parabola where   y = 4  and  b) the tangent lines when  x = −2.
By applying the implicit derivation on the parabola we get the slope of the tangent line:
2y dy/dx+3 dy/dx-2=0
dy/dx=2/(2y+3)
a)  Substitute the value   y = 4   to find the slope of the tangent line: m=dy/dx=2/(2∙2+3)=2/7
The   x1   coordinate of the tangent line corresponding to the value   y1 = 4   is:
x_1=(y^2+3y+4)/2=(4+6+4)/2=7
Now we can find the tangent line according to a point and the slope:       y = mx + (y1 − mx1)
y=1/3 x+4-2/3
And after arranging terms we get the tangent line equation 3y = x + 10
b) To find the  y  value of the point  x = −2   we must solve the equation     y22y −2x − 4 = 0
y22y = 0 y(y − 2) = 0
The solutions are  y1 = 0  and  y2 = 2  and the two tangent points are:  (−2 , 0)  and  (−2 , 2)
The first slope of tangent line is: m_1=dy/dx=1/(y-1)=1/(0-1)=-1
The second slope of tangent line is: m_2=dy/dx=1/(y-1)=1/(2-1)=1
The equation of first tangent line is: y = − 1∙x + (0 − 1 ∙ 2) = −x −2 y = −x −2
The equation of second tangent line is: y = 1∙ x + 2 − 1 ∙ (− 2) = x + 4 y = x + 4
Example 4a - tangent line Print parabola tangent lines example
Find the equations of the tangent lines to the parabola given by the equation   (y + 1)2 = x − 5
at the point where   x = 6.
First we find the corresponding  y  values at  x = 6:             (y + 1)2 = 6 − 5
Solving this equation we get:             y2 + 2y = 0           and the solutions are   y1 = 0   and   y2 =2
Now applying the implicit derivation on the parabola we get the slope   dy/dx   of the tangent line:
2(y+1)  dy/dx=1
dy/dx=-1/2(y+1)(1)
Substitute the values of   y   into equation  (1)   in order to find the slopes of the tangent lines:
For the value of   y = 0   we get dy/dx = m1 = 0.5
And for the value of   y = − 2   we get m2 =0.5
Now we can find the tangent line according to a point and the slope:       y = mx + (y1 − mx1)
For the first point (6 , 0)   The tangent line is: y = 0.5x − 0.5 ‧ 6 y = 0.5x − 3
For the second point (6 , −2)   The tangent line is: y =0.5x − 2 + 0.5 ‧ 6 y =0.5x + 1
sketch of example 4a The drawing of the parabola and the tangent lines see sketch at left
Example 4b - tangent line Print parabola tangent lines example
Find the equations of the tangent lines to the tilted parabola given by the equation
x2 + 4xy + 4y22x + 12y + 8 = 0   and the intersection points of the line x = 0 and the parabola.
First step is to find the coordinates of the line and parabola so we shell substitute the value of the line x = 0 into the parabola equation       y2 + 3y + 2 = 0
y_1,2=(-3±√(9-8))/2=-2 , -1
We found the two intersection points (0 , −1)   and   (0 , −2) and we have to find the tangent lines at those points.
To find the slope of the tangent line we shell explicitly derivate the equation of the parabola
2x+4y+4x dy/dx+8y dy/dx-2+12 dy/dx=0
dy/dx (4x+8y+12)+2x+4y-2=0
dy/dx=(x+2y-1)/(2x+4y+6)(1)
sketch of example 4b
The equation of a line defined by a point and the slope is:
y = mx + y1 − mx1
For point (0 , -1) m=(-2-1)/(-4+6)=-1.5
Tangent line: 1.5x y − 1 = 0
For point (0 , -2) m=(-4-1)/(-8+6)=2.5
Tangent line: 2.5x + y + 2 = 0
Example 5 - tilted parabola Print tilted parabola example
Find the equation and the vertex point of a parabola. If the focus point is at  (1 , −2)  and directrix at the line   x − 2y + 3 = 0.
Sketch ex 5
We choose a point  (x , y)  along the parabola, according to the parabola definition, distances are equal:
FE = ED
The length of the line  DE  can be found by the equation of the distance between a point and a line.
DE=|mx+ny+q|/√(m^2+n^2 )
The distance  FE  is:
FE=√((x-f_x )^2+(y-f_y )^2 )
Substituting values to DE  = FE |mx+ny+d|/√(m^2+n^2 )=√((x-f_x )^2+(y-f_y )^2 ) (1)
taking the square of both sides (mx+ny+d)^2/(m^2+n^2 )=(x-f_x )^2+(y-f_y )^2 (2)
After a few steps we get the final equation of a tilted parabola when the directrix and the focus are given.
n2 x2 − 2mnxy + m2y22(m2fx + n2fx + mq)x − 2(m2fy + n2fy + nq)y + (m2 + n2)(fx2 + fy2) − q2 = 0
Notice the term   −2mnxy   this term is a part of the tilted parabola.
Substitute the given values into eq  (2)  we get: (x-2y+3)^2/5=(x+1)^2+(y+2)^2
After multiplying and arranging terms we get the tilted parabola equation:
4x^2+4xy+y^2+4x+32y+16=0
The vertex is located half way between the focus and the point  G.
Point G is the intersection of the symmetry line and the directrix.
The symmetry line is passing through the focus and is perpendicular to the directrix line.
Slope of the directrix is: md = 0.5
The slope of the symmetry line is: m_s=-1/m_d =-1/0.5=-2
The equation of symmetry line: y = mx + y1 − mx1 = msx + yfocus − msxfocus
y =2x − 4
Aafter solving both equations the intersection point between symmetry line and directrix line is:       G(−2.2 , 0.4)
And the vertex is: (v_x  ,v_y )=((d_x+f_x)/2   ,(d_y+f_y)/2)=((-2.2-1)/2  ,(0.4-2)/2)=(-1.6 ,-0.8)
Example 5a - tilted parabola Print tiltrd parabola example
Find the equation of a parabola with focus at point  (4 , 1)  and vertex at point  (2 , 3) .
Sketch ex 5a
Figure 2
Figure 2
The slope of the symmetry line can be found by the slope of both points:
m=(f_y-v_y)/(f_x-v_x )
The slope of the directrix is:
M=-1/m=(v_x-f_x)/(f_y-v_y )
The coordinate of the intersection point of the symmetry line and the directrix line, based on the fect that   FV = VD   (see figure 2) is
[v_x-(f_x-v_x )  ,v_y+(v_y-f_y )]
D(2v_x-f_x  ,2v_y-f_y )
D(−44 , 6 − 1)      →      (−8 , 5)
The equation of the directrix line is according to point D and the slope  m:
y = Mx + dy − Mdx
y = 3x + 29     →     3x − y + 29 = 0
According to the parabola definition the distance     EG = FE
|Ax+By+C|/√(A^2+B^2 )=√((x-f_x )^2+(y-f_y )^2 )
|3x-y+29|/√(9+1)=√((x-4)^2+(y-1)^2 )
After taking the square values of both sides we get the parabola's equation
x2 + 6xy + 9y2254x + 38y − 671 = 0
Example 5b - tilted parabola Print tiltrd parabola example
Find the equation of a parabola with focus at point  (1.5 , 1)  and vertex at point  (1 , 3) .
Sketch ex 5a
Figure 2
Figure 2
The slope of the symmetry line can be found by the slope of both points:
m=(f_y-v_y)/(f_x-v_x )=(1-3)/(-1.5+1)=4
The slope of the directrix is:
M=-1/m=(v_x-f_x)/(f_y-v_y )=-0.25
The coordinate of the intersection point of the symmetry line and the directrix line, based on the fect that   FV = VD   (see figure 2) is
[v_x-(f_x-v_x )  ,v_y+(v_y-f_y )]
D(2v_x-f_x  ,2v_y-f_y )
D(−2 + 1.5 , 6 − 1)      →      (−0.5 , 5)
The equation of the directrix line is according to point D and the slope  m:
y = Mx + dy − Mdx
y =0.25x − 4.875   →   0.25x + y + 4.875 = 0
According to the parabola definition the distance     EG = FE
|Ax+By+C|/√(A^2+B^2 )=√((x-f_x )^2+(y-f_y )^2 )
|3x-y+29|/√(9+1)=√((x-4)^2+(y-1)^2 )
After taking the square values of both sides we get the parabola's equation
x2 + 6xy + 9y2254x + 38y − 671 = 0
Example 6 - Verify if a point is inside or outside of a parabola Print vertices and eccentricity example
Given a parabola with focus at point  (2 ,1)  and vertex at point  (4 , 1) determined if points
(1 , 3) and point (5 , − 2) are inside or outside the parabola.
Sketch ex 6
According to the definition of a parabola:
FB = BE
FB = √((x-f_x )^2+(y-f_y )^2 )
BE = |Ax+By+C|/√(A^2+B^2 )
Now we have to determine the values of A, B and C. So first we will find the point N.
Nx = (Vx − Fx) = 2Vx − Fx = 2 ‧ − 2 = 6
Ny = (Vy − Fy) = 2Vy − Fy = 2 ‧ 1 ‧ + 1 = 3
The slope of the symmetry line is:
m=(V_y-F_y)/(V_x-F_x )=(1+1)/(4-2)=1
The slope of the directrix is   M = −1
The directrix line equation according to point  N  and the slope  M  is:       Mx − y + y1 − Mx1 = 0
After substituting values we get − x − y + 3 + 6 = 0 x + y − 9 = 0
And the coefficients are:         A = 1         B = 1         C =9
Now we have to find the relations between the lengths FB and BE
If     FB = BE The point is on the parabola
If     FB < BE The point is inside the parabola
If     FB > BE The point is outside of the parabola
Check the first point: (1 , 3) BE=|1∙1+1∙3-9|/√(1+1)=5/√2=3.53
FB=√((1-2)^2+(3+1)^2 )=√(1+16)=√17=4.12
We see that:  FB > BE  so the point is outside of the parabola.
Check the first point: (5 , −2) BE=|1∙5+1∙(-2)-9|/√(1+1)=6/√2=4.24
FB=√((5-2)^2+(-2+1)^2 )=√(9+1)=√10=3.16
We see that:  FB < BE  so the point is inside the parabola.
Example 6a - Verify if a point is inside or outside of a parabola Print vertices and eccentricity example
Given the parabola  x22xy + y2 + 10x + 6y − 87 = 0 determined if the points (4 , 6) and the
point (6 , 1) are inside or outside of the parabola.
According to example 5b the value of q from eq (9) is:
q=(4F(C+A)-D^2-E^2)/(4D√C+4E√A)=(4∙(-87)2-100-36)/(4∙10+4∙6)=-13
Parabola of ex 6a
According to equations  (7)  and  (8)  the value of the focus coordinate is   (4 , 5)
f_x=(-2q√C-D)/2(A+C) =(2∙13-10)/(2∙2)=4
f_y=(-2q√A-E)/2(A+C) =(2∙13-6)/(2∙2)=5
According to equations  (14)  and  (15)  the value of the intersection of symmetry axis and the directrix is   (6 , 7)
d_x=(Af_x-q√C-√AC f_y)/(C+A)=(4+13-5)/2=6
d_y=(Cf_y-√AC f_x-q√A)/(C+A)=(5-4+13)/2=7
Now we can find the coordinate value of the vertex according to equations (16).
v_x=(d_x+f_x)/2=(6+4)/2=5 v_y=(d_y+f_y)/2=(7+5)/2=6
The slope of the parabola symmetry line is m=(v_y-f_y)/(v_x-f_x )=(6-5)/(5-4)=1
The slope of the directrix is: M=-1/m=-1
The directrix equation according to slope and point (dx , dy) is:           y = mx + (y1 − mx1)
y = − x + (7 + 6) x + y − 13 = 0
According to the parabola definition we have:     FP = PA   (where P is the point to verify).
If     FP = PA The point is on the parabola   (see example 6) 
If     FP < PA The point is inside the parabola
If     FP > PA The point is outside of the parabola
Check the first point: (4 , 6) FP=√((4-4)^2+(6-5)^2 )=1
PA=|4+6-13|/√(1+1)=3/√2=2.12
We see that:  PA > FP  so the point is inside the parabola.
Check the first point: (6 , 1) FP=√((6-4)^2+(1-5)^2 )=√20=4.47
PA=|6+1-13|/√(1+1)=6/√2=4.24
We see that:  PA < FP  so the point is outside of the parabola.
Example 6b - Verify if a point is inside or outside of a parabola Print vertices and eccentricity example
Given the parabola  y24y − 4x + 8 = 0 determined if the points (1 , 0) and (2 , 0) and (3 , 0)
are inside or outside of the parabola.
Parabola of ex 6b
From the equation of the parabola we can verify that the parabola is horizontal  (y2).
We will solve the problem by the general case of tilted parabola as described in example 5b in this case the coefficients values are:
A = 0 B = 0 C = 1 D =4 E =4 F = 8
q=(4F(C+A)-D^2-E^2)/(4D√C+4E√A)=(4∙8(1)-16-16)/(-4∙4+0)=0
According to equations  (7)  and  (8)  the value of the focus coordinate is   (4 , 5)
f_x=(-2q√C-D)/2(A+C) =(-2∙0+4)/(2∙1)=2
f_y=(-2q√A-E)/2(A+C) =(-2∙0+4)/(2∙1)=2
According to equations  (14)  and  (15)  the value of the intersection of symmetry axis and the directrix is   (6 , 7)
d_x=(Af_x-q√C-√AC f_y)/(C+A)=(0+0-0)/1=0
d_y=(Cf_y-√AC f_x-q√A)/(C+A)=(1∙2)/1=2
The slope of the parabola symmetry line is m=(f_y-d_y)/(f_x-d_x )=(2-2)/(2-0)=0
The slope of the directrix is: M=-1/m=-∞   (vertical)
The directrix equation is passing through point (0,2) so the equation of the directrix is    x = 0
if line equation is:   mx + ny + q = 0 then     m = 1       n = 0       q = 0
According to the parabola definition we have:     FP = PA   (where P is the point to verify).
If     FP = PA The point is on the parabola   (see example 6) 
If     FP < PA The point is inside the parabola
If     FP > PA The point is outside of the parabola
Check the first point: (1 , 0) FP=√((x-f_x )^2+(y-f_y )^2 )=√((1-2)^2+(0-2)^2 )=√5=2.23
PA=|mx+ny+q|/√(m^2+n^2 )=|1∙1+0+0|/√(1+0)=1/1=1
We see that:  FP > PA  so the point is outside of the parabola.
Check the first point: (2 , 0) FP=√((2-2)^2+(0-2)^2 )=2
PA=|1∙2+0+0|/√(1+0)=2/1=2
We see that:  FP = PA  so the point is on the parabola.
Check the first point: (3 , 0) FP=√((3-2)^2+(0-2)^2 )=√5=2.23
PA=|1∙3+0+0|/√(1+0)=3/1=3
We see that:  FP < PA  so the point is inside the parabola.