Parabola calculator Print parabola calculator
( x )2 = ( y )
x2 + x + y + = 0
Parabola figure - 1
Focus
( , )
Vertex
( , )
Directrix equation
Symmetry axis
Input limit:
Notes:
                 
Parabola equation:
Parabola and line intersection
y = x +
Intersection: Point 1
Point 2
Parabola with center at (0,0) Inclined parabola Parabola tangent line ex: 4 , 4a
Parabola tables Foci vertex ex: 1 , 1a , 1b , 1c Inclined parabola ex: 5 , 5a
Parabola tangent lines Parabola equation ex: 2 , 2a
Parabola and line intersection Parabola and line ex: 3 , 3a
Parabola - with center at  (0 , 0)  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.
Vertex at (0,0) summary           p > 0 Print hyperbola summary
Type Sketch parabola equation Focus at
Vertex
Directrix equation
1 Parabola figure - 3 x2 = 2py
(0 , p/2)
(0 , 0)
y = −p/2
2 Parabola figure - 4 x2 = −2py
(0 , − p/2)
(0 , 0)
y = p/2
3 Parabola figure - 1 y2 = 2px
(p/2 , 0)
(0 , 0)
x = −p/2
4 Parabola figure - 2 y2 = −2px
(−p/2 , 0)
(0 , 0)
x = p/2
Vertex at (a,b) summary           p > 0 Print hyperbola summary
Type sketch parabola equation Focus at
Vertex
Directrix equation
1 Parabola figure - 3 (x − a)2 = 2p(y − b)
(a , b + p/2)
(a , b - p/2)
y = b − p/2
2 Parabola figure - 4 (x − a)2 = −2p(y − b)
(a , b − p/2)
(a , b + p/2)
y = b + p/2
3 Parabola figure - 1 (y − b)2 = 2p(x − a)
(a + p/2 , b)
(b - p/2 , b)
x = a − p/2
4 Parabola figure - 2 (y − b)2 = −2p(x − a)
(a − p/2 , b)
(b + p/2 , b)
x = a + p/2
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

Parabola tangent line
The general form of a horizontal parabola is.
(y − b)2 = 2p(x − a)
By implicit diferentiation 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
Parabola and line intersection points Print parabola and line intersection summary
Parabola and line intersection
If the parabola and the line are given by the equations:
αx2 + βx + γy + δ = 0(1)
Ax + By + C = 0(2)
Solving this equations we get the intersection points.
From equation (2) we have
y=(-Ax-C)/B(3)
Substitute the value of  y  into eq.  (1)  to get  x.
αx^2+βx-γ((Ax+C)/B)+δ=0
After multiplying all terms by B we have:αBx2 + (βB − γA)x − γC + δB = 0
And  x  is equal to:x=(γA-βB±√((βB-γA)^2+4αB(δB-γC) ))/2αB
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
Intersections Condition
Two points (βB − γA)2 + 4αB(δB − γC) > 0
One point (tangent) (βB − γA)2 + 4αB(δB − γC) = 0
No intersection (βB − γA)2 + 4αB(δB − γC) < 0
Parabola and line intersection If the parabola is horizontal then the equation of the parabola is:
αy2 + βy + γx + δ = 0(4)
Ax + By + C = 0(5)
Solving this equations we get the intersection points.
From equation  (5)  we have
x=(-By-C)/A(6)
Substitute the value of  x  to eq.  (4)  to get  y.
y=(-Ax-C)/B(7)
After multiplying all terms by  B  we get:αAy2 + (βA − γB)y − γC + δA = 0
And  y  is equal to:y_1,2=(γB-βA±√((βA-γB)^2-4αA(δA-γC) ))/2αA
The value of  x  is found according to equation  (6)
The conditions for intersection are:
Intersections Condition
Two points (βA − γB)2 4αA(δA − γC) > 0
One point (tangent) (βA − γB)2 4αA(δA − γC) = 0
No intersection (βA − γB)2 4αA(δA − γC) < 0
Inclined parabola Print inclined parabola summary
Inclined parabola
According to the parabola definition we have
FP = PA
FP=√((x-F_x )^2+(y-F_y )^2 )
From the equation of the distance from point to line we have the expression:
PA=|Ax+By+C|/√(A^2+B^2 )
√((x-F_x )^2+(y-F_y )^2 )=|Ax+By+C|/√(A^2+B^2 )
After multiplying 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).
B^2 x^2-2ABxy+A^2 y^2-(2AC+2f_x )x+(2f_y-2BC)y+(A^2+B^2 )(〖f_x〗^2+〖f_y〗^2 )=0
Example 1 - vertices and foci Print vertices and foci example
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 that is 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 at 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 oposite 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 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 openning of the parabola is to the left.
The symetry axis is at  y = 2   (see sketch)
The directrix and the symetry 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 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 + d
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 line equation into the parabola equation: x2 + ( A + Bm)x + Bd + 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 line   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 it's openning is to the right direction (see the 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 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 5 - tilted parabola Print vertices and eccentricity example
Find the equation of a parabola with focus at point  (1 , −2)  and directrix at 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.
d=|Ap_x+Bp_y+C|/√(A^2+B^2 )
The distance  FE  is:
FE=√((x-f_x )^2+(y-f_y )^2 )
Substituting values to FE  = DE |Ap_x+Bp_y+C|/√(A^2+B^2 )=√((x-f_x )^2+(y-f_y )^2 ) (1)
taking the square of both sides (Ap_x+Bp_y+C)^2/(A^2+B^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.
B^2 x^2-2ABxy+A^2 y^2-(2AC+2f_x )x+(2f_y-2BC)y+(A^2+B^2 )(〖f_x〗^2+〖f_y〗^2 )=0
Notice the term   −2ABxy   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
Example 5a - tilted parabola Print vertices and eccentricity example
Find the equation of a parabola with focus at point  (4 , 1)  and vertex at point  (2 , 3) .
Sketch ex 5
The slope of the symmetry line can be found by the tangent 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 of symmetry line and the directrix is
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:
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