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|

We denote the distance from the focus to the origin as

p, then from the distance formula we have |FP = PA|

After squaring both sides we get:

And finally we get the general equation of a parabola:

x^{2} − 2py = 0

notice that p > 0

Notice that this parabola has the vertex at the origin and focus distance of p from the directrix.

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).

Find the coordinate of the foci and the directrix equation of the parabola given by the equation y^{2}= −12x.

We mark the distance between the directrix and the focus by p then y^{2}= −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:

The focus of a parabola is located at the point (0 , 2) and the vertex at (0 , 4) find the equation of the parabola.

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:

d_{1}= d_{2}

After squaring both sides we get

(x − 0)^{2}+ (y − 2)^{2}= (y − 6)^{2}

x^{2}+ y^{2} − 4y + 4 = y^{2} − 12y + 36

After arranging terms we get the equation of the parabola:

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.

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).

Find the intersection points of a parabola (y − 1)^{2} − 2(x + 2.5) = 0 and the line a) x = 0 and b) the line y = 0.

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.

y^{2} − 2y − 2x − 4 = 0

a) Substitute line equation x = 0 into the parabola equation we get.

y^{2} − 2y − 4 = 0

The solutions of this equation are:

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)

A parabola is given by y^{2} −2y −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:

a) Substitute the value y = 4 to find the slope of the tangent line:

The x_{1} coordinate of the tangent line corresponding to the value y_{1} = 4 is:

Now we can find the tangent line according to a point and the slope: y = mx + (y_{1} − mx_{1})

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 y^{2} − 2y −2x − 4 = 0

y^{2} − 2y = 0

⟶

y(y − 2) = 0

The solutions are y_{1}= 0 and y_{2}= 2 and the two tangent points are: (−2 , 0) and (−2 , 2)