Question

Let L be a normal to the parabola $ {y^2} = 4x. $ if L passes through the point $ \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 \\
  $

Answer 1

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 - {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 $ {y^2} = 4x. $
Given point is $ \left( {9,6} \right) $
Substituting above point in given equation of parabola:
We have,
 $
  {\left( 6 \right)^2} = 4\left( 9 \right) \\
   \Rightarrow 36 = 36 \;
  $
We clearly see that the given point $ \left( {9,6} \right) $ satisfies the given equation of parabola.
Therefore, given point $ \left( {9,6} \right) $ lies on parabola.
To find the equation of normal to parabola we first find the equation of tangent at point $ \left( {9,6} \right) $ .
For this we first find the derivative of a given equation of parabola. We have,
 $
  \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 $ \left( {9,6} \right) $ in above we have,
 $
  \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 $ \left( {9,6} \right) $ is $ \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 $ and normal passes through point $ \left( {9,6} \right) $ .
Therefore, equation of normal will be given as:
 $ y - {y_1} = m\left( {x - {x_1}} \right) $
Substituting values in above equation. We have,
 $
  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 = 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 - a{m^3} $ . Since it passes through point $ \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.