Question

The product of the slopes of the non-horizontal normals drawn through the point \[ (6,0) \] to the parabola \[ y^2=8x \] is:

• A. \(1\)
• B. \(-2\)
• C. \(2\)
• D. \(-1\)

Hint:

For the parabola \[ y^2=4ax, \] the normal at parameter \(t\) is \[ y=-tx+2at+at^3. \] This formula is the quickest way to solve normal-related questions in competitive examinations.

Answer 1

The Correct Option is D

Step 1: Set up the parabola.
The parabola is $y^2=8x$. Comparing with $y^2=4ax$, we get $4a=8$, so $a=2$.

Step 2: Write the normal in slope form.
The normal to $y^2=4ax$ with slope $m$ is \[ y=mx-2am-am^3. \] With $a=2$ this is $y=mx-4m-2m^3$.

Step 3: Make it pass through the given point.
The normal must pass through $(6,0)$. Put $x=6,\,y=0$: \[ 0=6m-4m-2m^3. \]

Step 4: Simplify the equation in $m$.
\[ 0=2m-2m^3=2m(1-m^2). \] So $m=0$ or $m^2=1$, giving $m=0,\,1,\,-1$.

Step 5: Keep only non-horizontal normals.
A horizontal normal has slope $0$, so we drop $m=0$. The non-horizontal slopes are $m=1$ and $m=-1$.

Step 6: Multiply the remaining slopes.
\[ (1)\times(-1)=-1. \] So the product of the slopes is $-1$, which is option 4.
\[ \boxed{-1} \]