Question

The line \[x + y = 6\] is a normal to the parabola \[{y^2} = 8x\] at the point
A. \[(8, -12)\]
B. \[(4,2)\]
C. \[(2,4)\]
D. \[(8,8)\]

Answer 1

Hint: Using the equation of line, we find the slope of the line. Comparing the equation of parabola to the general equation of parabola we find the value of a. Substitute the values in the general point in terms of slope which gives the equation of normal.
* General equation of the line is \[y = mx + c\] where m is the slope of the line.
* General equation of parabola is \[{y^2} = 4ax\]
* At point \[(a{m^2}, - 2am)\], the equation of normal to the parabola with slope m and constant a is given by \[y = mx - 2am - a{m^3}\].

Complete step-by-step answer:
We have an equation of line as \[x + y = 6\].
We can write the equation of line by shifting the variable x to RHS of the equation
\[y = - x + 6\]
Comparing the equation to the general equation of line i.e. \[y = mx + c\].
We get the value of slope as \[m = - 1\] … (1)
Now we are given an equation of the parabola as \[{y^2} = 8x\].
Comparing the equation to the general equation of parabola i.e. \[{y^2} = 4ax\].
We get the value of \[a = 2\]. … (2)
Now we know that equation of normal to the parabola at the point \[(a{m^2}, - 2am)\] is given by \[y = mx - 2am - a{m^3}\].
So, we can find the point by substituting the values of a and m from equations (1) and (2) in \[(a{m^2}, - 2am)\].
\[
   \Rightarrow (a{m^2}, - 2am) = \left( {(2){{( - 1)}^2}, - 2(2)( - 1)} \right) \\
   \Rightarrow (a{m^2}, - 2am) = \left( {2 \times 1,2 \times 2} \right) \\
   \Rightarrow (a{m^2}, - 2am) = \left( {2,4} \right) \\
 \]
Therefore, the line \[x + y = 6\] is a normal to the parabola \[{y^2} = 8ax\] at the point \[(2,4)\].

So, the correct answer is “Option C”.

Note: Students make the mistake of solving this question using the general equation of normal to a parabola which is in terms of point which is wrong, because then we will be needing two points and here we don’t have any point.