Question

If \(2x+y+k=0\) is a normal to the parabola \(y^2=-8x\), then the value of \(k\) is

• A. 8
• B. 16
• C. 24
• D. 32

Hint:

For \(y^2=4ax\), parametric point is \((at^2,2at)\) and normal slope is \(-t\). Match slope with given normal and substitute point to find constant.

Answer 1

The Correct Option is C

Step 1: Write parabola in standard form.
Given parabola:
\[ y^2=-8x \] Compare with \(y^2=4ax\):
\[ 4a=-8 \Rightarrow a=-2 \] Step 2: Parametric point on parabola.
For \(y^2=4ax\), parametric point is \((at^2,2at)\).
So here:
\[ x=at^2=-2t^2,\quad y=2at=-4t \] Step 3: Slope of normal for standard parabola.
For parabola \(y^2=4ax\), slope of tangent is \(\frac{1}{t}\).
So slope of normal is \(-t\).
Step 4: Given normal line and its slope.
Given line:
\[ 2x+y+k=0 \Rightarrow y=-2x-k \] So slope = \(-2\).
Thus:
\[ -t=-2 \Rightarrow t=2 \] Step 5: Find the point on parabola.
Substitute \(t=2\):
\[ x=-2(2^2)=-8,\quad y=-4(2)=-8 \] Step 6: Use point on the line to find \(k\).
Since \((-8,-8)\) lies on \(2x+y+k=0\):
\[ 2(-8)+(-8)+k=0 \Rightarrow -16-8+k=0 \Rightarrow k=24 \] Final Answer: \[ \boxed{24} \]