$ { ext{Write an equation of a horizontal parabola with a vertex of }}\left( {1, - 2} ight){ ext{ and passing }} \ { ext{through }}\left( {9, - 4} ight). \ $

Question

$  {	ext{Write an equation of a horizontal parabola with a vertex of }}\left( {1, - 2} ight){	ext{ and passing }}   \  {	ext{through }}\left( {9, - 4} ight).   \ $

Vedantu Content Team · Accepted Answer

$  {	ext{Let the equation of the horizontal parabola is }}{\left( {y - h} ight)^2} = 4a\left( {x - k} ight)   \  {	ext{because it is passing through}}\left( {9, - 4} ight)\;{	ext{which is right side of the parabola }}   \  {	ext{i}}{	ext{.e parabola is opening right side}}{	ext{.}}   \  {	ext{so the vertex of the parabola is }}\left( {k,h} ight)   \  {	ext{So, compare the vertex from the given vertex}}   \  {	ext{i}}{	ext{.e }}\left( {1, - 2} ight)   \   \Rightarrow k = 1,\;h =  - 2   \  {	ext{So, the equation of parabola is}}   \  {\left( {y + 2} ight)^2} = 4a\left( {x - 1} ight)   \  {	ext{Now it is passing through }}\left( {9, - 4} ight){	ext{ which satisfy the parabola}}   \   \Rightarrow {\left( { - 4 + 2} ight)^2} = 4a\left( {9 - 1} ight)   \   \Rightarrow 4 = 4a 	imes 8 \Rightarrow a = \dfrac{1}{8} \Rightarrow 4a = \dfrac{1}{2}   \   \Rightarrow {	ext{Equation of parabola is}}   \  {\left( {y + 2} ight)^2} = \dfrac{1}{2}\left( {x - 1} ight)   \  {	ext{So, this is your required equation}}{	ext{.}}   \  {	ext{NOTE: -  In this type of problem first check from given condition parabola is opening which side,}}   \  {	ext{              then satisfying the condition according to given problem statement you will get your answer}}{	ext{.}}   \ $

$
{\text{Write an equation of a horizontal parabola with a vertex of }}\left( {1, - 2} \right){\text{ and passing }} \\
{\text{through }}\left( {9, - 4} \right). \\
$

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$ {\text{Write an equation of a horizontal parabola with a vertex of }}\left( {1, - 2} \right){\text{ and passing }} \\ {\text{through }}\left( {9, - 4} \right). \\ $

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