# What is the equation of the parabola that has a vertex at  (21, 11)  and passes through point  (23,-4) ?

Jun 14, 2018

$2 {\left(y - 11\right)}^{2} = 225 \left(x - 21\right)$ (Parabola opened rightwards,(i.e,)towards positive x direction)

#### Explanation:

The General equation of a parabola is ${\left(y - k\right)}^{2} = 4 a \left(x - h\right)$

(Parabola opened towards positive x-direction)

where

$a$ is a arbitrary constant,

( $h , k$) is the vertex.

Here we have our vertex as ( $21 , 11$).

SUBSTITUTE the x and y coordinate values of the vertex in the equation above, we get.

${\left(y - 11\right)}^{2} = 4 a \left(x - 21\right)$

In order to find the value of ' $a$' substitute the given point in the equation

then we get

${\left(- 4 - 11\right)}^{2} = 4 a \left(23 - 21\right)$
$\implies {\left(- 15\right)}^{2} = 8 a$

$\implies a = \frac{225}{8}$

Substitute the value for ' $a$' In the above equation to have the equation of the required parabola.

${\left(y - 11\right)}^{2} = 4 \cdot \frac{225}{8} \left(x - 21\right)$
$\implies 2 {\left(y - 11\right)}^{2} = 225 \left(x - 21\right)$

$\textcolor{b l u e}{N O T E} :$

The general equation of a parabola "OPENED UPWARDS " will

results in a slightly different equation , And leads to a different

answer . Its general form will be

${\left(x - h\right)}^{2} = 4 \cdot a \left(y - k\right)$

where (h,k) is the vertex..,