Question

What is the equation of the parabola that has a vertex at `(-1, 7)` and passes through point `(2,-3)`?

Answer 1

If the axis is assumed to be parallel to the x-axis, `(y-7)^2=100/3(x+1)` See explanation for the equation of the family of parabolas, when there is no such assumption.

Explanation:

Let the equation of axis of the parabola with vertex `V(-1, 7)` be

`y-7=m(x+1)`, with m not equal tom 0 nor `oo`..

Then the equation of the tangent at the vertex will be

`y-7=(-1/m)(x+1)`.

Now, the equation of any parabola having V as vertex is

`(y-7-m(x+1))^2= 4a(y-7+(1/m)(x+1))`.

This passes through `(2, -3)`, if

`(-10-3m)^2=4a(3/m-10)`. This gives the relation between the two

parameters a and m as

`9m^3+60m^2+(100+40a)m-12a=0`.

In particular, if the axis is assumed to be parallel to the x-axis, m = 0,

this method can be ignored.

In this case, `y-7=0` is for the axis and x+1 = 0 is for the tangent at

the vertex. and the equation of the parabola becomes

`(y-7)^2=4a(x+1).`

As it passes through (2, -3), a =25/3.

The parabola is given by

`(y-7)^2=100/3(x+1)`

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