Question

If the perpendicular distance from the focus of a parabola \(y^2=4ax\) to its directrix is \( \frac{3}{2} \), then the equation of the normal drawn at \( (4a, -4a) \) is

• A. \( 2x+y=3 \)
• B. \( 2x-y=9 \)
• C. \( x-2y=9 \)
• D. \( x+2y+3=0 \)

Hint:

For parabola \(y^2=4ax\): Focus is \((a,0)\), directrix is \(x=-a\). Distance between focus and directrix is \(2a\). Equation of normal at \( (x_1,y_1) \) is \( y-y_1 = -\frac{y_1}{2a}(x-x_1) \). Alternatively, the normal at \( (at^2, 2at) \) is \( y+tx = 2at+at^3 \). The point \( (4a, -4a) \) corresponds to \( at^2 = 4a \implies t^2=4 \implies t=\pm 2 \). And \( 2at = -4a \implies t = -2 \). So \( t=-2 \). Normal: \( y+(-2)x = 2a(-2)+a(-2)^3 \implies y-2x = -4a -8a = -12a \). \( 2x-y=12a \). Same result.

Answer 1

The Correct Option is B

For the parabola \(y^2=4ax\): Focus is \( S=(a,0) \).
Equation of directrix is \( x = -a \) or \( x+a=0 \).
The perpendicular distance from the focus \( (a,0) \) to the directrix \( x+a=0 \) is: \[ \frac{|a+a|}{\sqrt{1^2+0^2}} = \frac{|2a|}{1} = |2a| \] Given this distance is \( \frac{3}{2} \).
So, \( |2a| = \frac{3}{2} \).
This means \( 2a = \frac{3}{2} \) or \( 2a = -\frac{3}{2} \).
Typically, for \(y^2=4ax\), \(a\) is considered positive, so \( 2a = \frac{3}{2} \implies a = \frac{3}{4} \).
The equation of the normal to \(y^2=4ax\) at a point \( (x_1, y_1) \) is \( y-y_1 = -\frac{y_1}{2a}(x-x_1) \).
The point is \( (4a, -4a) \).
So \( x_1 = 4a, y_1 = -4a \).
Substitute these into the normal equation: \[ y - (-4a) = -\frac{-4a}{2a}(x - 4a) \] \[ y + 4a = -(-2)(x - 4a) \] \[ y + 4a = 2(x - 4a) \] \[ y + 4a = 2x - 8a \] \[ 2x - y - 12a = 0 \] Substitute \( a = 3/4 \): \[ 2x - y - 12\left(\frac{3}{4}\right) = 0 \] \[ 2x - y - 3 \times 3 = 0 \] \[ 2x - y - 9 = 0 \] \[ 2x - y = 9 \] This matches option (2).