Question

The equation of a normal to the parabola $y^2 = 4 x$, which passes through the point (9, 6) is :

• A. $x + 3y + 15 = 0$
• B. $2x - y + 12 = 0$
• C. $2x + y + 12 = 0$
• D. $2x - y - 12 = 0$

Hint:

To verify the answer quickly, test the coordinates of the given point $(9, 6)$ in the options:
- For Option (D): $2(9) - 6 - 12 = 18 - 18 = 0$. This coordinates verification is extremely fast.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The normal to a parabola can be expressed in slope form. For the standard parabola $y^2 = 4ax$, any normal line has a specific mathematical relation depending on its slope $m$.
Key Formula or Approach:
The equation of a normal to the parabola $y^2 = 4ax$ in slope form is:
\[ y = mx - 2am - am^3 \]
Here, for $y^2 = 4x$, we have $a = 1$.

Step 2: Detailed Explanation:
Substitute $a = 1$ into the normal equation:
\[ y = mx - 2m - m^3 \]
Since the normal passes through the point $(9, 6)$, substitute $x = 9$ and $y = 6$ into this equation:
\[ 6 = 9m - 2m - m^3 \]
\[ m^3 - 7m + 6 = 0 \]
Let us find the roots of this cubic equation.
By inspection, $m = 1$ is a root because $1^3 - 7(1) + 6 = 0$.
We can factor out $(m - 1)$:
\[ (m - 1)(m^2 + m - 6) = 0 \]
Factor the remaining quadratic expression:
\[ (m - 1)(m + 3)(m - 2) = 0 \]
Thus, the three possible slopes are $m = 1$, $m = 2$, and $m = -3$.
Let us write the equations of the normals corresponding to these slopes:
1. For $m = 1$:
\[ y = x - 2 - 1 \implies x - y - 3 = 0 \]
2. For $m = 2$:
\[ y = 2x - 4 - 8 \implies 2x - y - 12 = 0 \]
3. For $m = -3$:
\[ y = -3x + 6 + 27 \implies 3x + y - 33 = 0 \]
Comparing with the given choices, the equation $2x - y - 12 = 0$ is present as Option (D).
Therefore, the correct equation is $2x - y - 12 = 0$.

Step 3: Final Answer
The correct option is (D).