Answer
Verified
37.5k+ views
$
{\text{As you know that general equation of a horizontal parabola is:}} \\
{\left( {y - k} \right)^2} = 4a\left( {x - h} \right){\text{ }} \\
{\text{As Vertex is given to us as }}\left( { - 8,2} \right){\text{, therefore we have}} \\
{\left( {y - 2} \right)^2} = 4a\left( {x - \left( { - 8} \right)} \right){\text{ }} \\
\Rightarrow {\left( {y - 2} \right)^2} = 4a\left( {x + 8} \right){\text{ }} - Equation(1){\text{ }} \\
{\text{Now, putting }}\left( {2, - 3} \right){\text{ in Equation(1) we get}} \\
{\left( { - 3 - 2} \right)^2}{\text{ }} = {\text{ }}4a\left( {2 + 8} \right) \\
\Rightarrow {\left( { - 5} \right)^2}{\text{ }} = {\text{ }}4a\left( {10} \right){\text{ }} \\
\Rightarrow 25 = 40a \\
\Rightarrow a = \frac{5}{8}{\text{ }} - Equation(2) \\
{\text{Using Equation(2) in Equation(1)}} \\
\Rightarrow {\left( {y - 2} \right)^2} = 4\left( {\frac{5}{8}} \right)\left( {x + 8} \right) \\
\Rightarrow {\left( {y - 2} \right)^2} = \frac{5}{2}\left( {x + 8} \right){\text{ is the required Equation}} \\
{\text{Note: For these kinds of questions we must remember the general equation of a horizontal }} \\
{\text{parabola and then put vertex's coordinates in it to get the equation in only one variable 'a'}}{\text{. }} \\
{\text{Now, Put the coordinates if the point given to be on the parabola and solve to get value of 'a'}}{\text{. }} \\
{\text{Put the value of 'a' in the equation to get desired equation}}{\text{.}} \\
$
{\text{As you know that general equation of a horizontal parabola is:}} \\
{\left( {y - k} \right)^2} = 4a\left( {x - h} \right){\text{ }} \\
{\text{As Vertex is given to us as }}\left( { - 8,2} \right){\text{, therefore we have}} \\
{\left( {y - 2} \right)^2} = 4a\left( {x - \left( { - 8} \right)} \right){\text{ }} \\
\Rightarrow {\left( {y - 2} \right)^2} = 4a\left( {x + 8} \right){\text{ }} - Equation(1){\text{ }} \\
{\text{Now, putting }}\left( {2, - 3} \right){\text{ in Equation(1) we get}} \\
{\left( { - 3 - 2} \right)^2}{\text{ }} = {\text{ }}4a\left( {2 + 8} \right) \\
\Rightarrow {\left( { - 5} \right)^2}{\text{ }} = {\text{ }}4a\left( {10} \right){\text{ }} \\
\Rightarrow 25 = 40a \\
\Rightarrow a = \frac{5}{8}{\text{ }} - Equation(2) \\
{\text{Using Equation(2) in Equation(1)}} \\
\Rightarrow {\left( {y - 2} \right)^2} = 4\left( {\frac{5}{8}} \right)\left( {x + 8} \right) \\
\Rightarrow {\left( {y - 2} \right)^2} = \frac{5}{2}\left( {x + 8} \right){\text{ is the required Equation}} \\
{\text{Note: For these kinds of questions we must remember the general equation of a horizontal }} \\
{\text{parabola and then put vertex's coordinates in it to get the equation in only one variable 'a'}}{\text{. }} \\
{\text{Now, Put the coordinates if the point given to be on the parabola and solve to get value of 'a'}}{\text{. }} \\
{\text{Put the value of 'a' in the equation to get desired equation}}{\text{.}} \\
$
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
Other Pages
A convex lens is dipped in a liquid whose refractive class 12 physics JEE_Main
Identify which of the above shown graphs represent class 12 physics JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main