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

##### 1 Answer
Oct 4, 2016

If the axis is assumed to be parallel to the x-axis, ${\left(y - 7\right)}^{2} = \frac{100}{3} \left(x + 1\right)$ 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 $V \left(- 1 , 7\right)$ be

$y - 7 = m \left(x + 1\right)$, with m not equal tom 0 nor $\infty$..

Then the equation of the tangent at the vertex will be

$y - 7 = \left(- \frac{1}{m}\right) \left(x + 1\right)$.

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

${\left(y - 7 - m \left(x + 1\right)\right)}^{2} = 4 a \left(y - 7 + \left(\frac{1}{m}\right) \left(x + 1\right)\right)$.

This passes through $\left(2 , - 3\right)$, if

${\left(- 10 - 3 m\right)}^{2} = 4 a \left(\frac{3}{m} - 10\right)$. This gives the relation between the two

parameters a and m as

$9 {m}^{3} + 60 {m}^{2} + \left(100 + 40 a\right) m - 12 a = 0$.

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

${\left(y - 7\right)}^{2} = 4 a \left(x + 1\right) .$

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

The parabola is given by

${\left(y - 7\right)}^{2} = \frac{100}{3} \left(x + 1\right)$

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