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Question: Let L be a normal to the parabola \( {y^2} = 4x. \) if L passes through the point \( \left( {9,6} \r...

Let L be a normal to the parabola y2=4x.{y^2} = 4x. if L passes through the point (9,6)\left( {9,6} \right) , then L is given by:
A.  y−x+3=0 B.  y+3x−33=0 C.  y+x−15=0 D.  y−2x+12=0  A.\,\,y - x + 3 = 0 \\\ B.\,\,y + 3x - 33 = 0 \\\ C.\,\,y + x - 15 = 0 \\\ D.\,\,y - 2x + 12 = 0 \\\

Explanation

Solution

Hint : For this we first check whether a given point satisfies the equation of parabola and then find slope of tangent at this point then finding slope of normal using slope of a tangent and then finally using point and slope write equation of normal and solution of given problem.
Formulas used: Equation of a line in slope form: y−y1=m(x−x1),   Equation  of  normal  y= mx−2am−am3y - {y_1} = m\left( {x - {x_1}} \right),\,\,\,Equation\,\,of\,\,normal\,\,y = \,mx - 2am - a{m^3} .

Complete step-by-step answer :
Given equation of parabola is y2=4x.{y^2} = 4x.
Given point is (9,6)\left( {9,6} \right)
Substituting above point in given equation of parabola:
We have,
(6)2=4(9) ⇒36=36   {\left( 6 \right)^2} = 4\left( 9 \right) \\\ \Rightarrow 36 = 36 \;
We clearly see that the given point (9,6)\left( {9,6} \right) satisfies the given equation of parabola.
Therefore, given point (9,6)\left( {9,6} \right) lies on parabola.
To find the equation of normal to parabola we first find the equation of tangent at point (9,6)\left( {9,6} \right) .
For this we first find the derivative of a given equation of parabola. We have,
ddx(y)2=ddx(4x) ⇒2ydydx=4(1) ⇒dydx=42y ⇒dydx=2y  \dfrac{d}{{dx}}{\left( y \right)^2} = \dfrac{d}{{dx}}\left( {4x} \right) \\\ \Rightarrow 2y\dfrac{{dy}}{{dx}} = 4(1) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{2y}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{y} \\\
Substituting point (9,6)\left( {9,6} \right) in above we have,
dydx=26 ⇒dydx=13  \dfrac{{dy}}{{dx}} = \dfrac{2}{6} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{3} \\\
Hence, from above we can say that the slope of the tangent or slope of a curve at (9,6)\left( {9,6} \right) is 13\dfrac{1}{3} .
Since, normal is perpendicular to tangent.
Therefore its slope will be negative or reciprocal of slope of tangent.
Hence, slope of normal is = −3- 3 and normal passes through point (9,6)\left( {9,6} \right) .
Therefore, equation of normal will be given as:
y−y1=m(x−x1)y - {y_1} = m\left( {x - {x_1}} \right)
Substituting values in above equation. We have,
y−6=−3(x−9) ⇒y−6=−3x+27 ⇒y+3x=33   y - 6 = - 3\left( {x - 9} \right) \\\ \Rightarrow y - 6 = - 3x + 27 \\\ \Rightarrow y + 3x = 33 \;
Therefore, required equation of normal is y+3x−33=0y + 3x - 33 = 0
So, the correct answer is “Option B”.

Note : For this type of problem we can also find a solution to a given problem in another way. In this we use the standard equation of normal to give parabola. Which is written as y=mx−2am−am3y = mx - 2am - a{m^3} . Since it passes through point (9,16)\left( {9,16} \right) . Therefore, substituting this point and value of ‘a’ from parabola in above equation of normal to find value of ‘m’ and so equations of normal.