Question

If the normal drawn at P(8,16) to the parabola \( y^2=32x \) meets the parabola again at Q, then the equation of the tangent drawn at Q to the parabola is

• A. \( x+3y+72=0 \)
• B. \( x-y-120=0 \)
• C. \( 3x-y-264=0 \)
• D. \( x+y-24=0 \)

Hint:

The slope of the tangent at parameter \( t \) is \( 1/t \). Here slope at \( t=-3 \) is \( -1/3 \), matching the line equation \( x+3y... \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the parametric properties of the parabola. If normal at \( t_1 \) meets the curve at \( t_2 \), there is a specific relation between \( t_1 \) and \( t_2 \). Find \( t_2 \), get point Q, and find the tangent equation.
Step 2: Detailed Explanation:
Parabola \( y^2 = 32x \implies 4a = 32 \implies a = 8 \).
Point P(8,16) corresponds to \( (at^2, 2at) \).
\( 2(8)t_1 = 16 \implies t_1 = 1 \).
Relation for normal intersection: \( t_2 = -t_1 - \frac{2}{t_1} \).
\( t_2 = -1 - \frac{2}{1} = -3 \).
Point Q is \( (a t_2^2, 2a t_2) \):
\( x = 8(-3)^2 = 72 \).
\( y = 2(8)(-3) = -48 \).
Q is \( (72, -48) \).
Equation of tangent at \( Q(x_1, y_1) \): \( yy_1 = 2a(x+x_1) \).
\( y(-48) = 16(x + 72) \)
Divide by 16:
\( -3y = x + 72 \)
\( x + 3y + 72 = 0 \).

Step 3: Final Answer:
The equation is \( x+3y+72=0 \).