Question

A tangent is drawn to the parabola $y^2=4x$ at a point $P$ on the parabola in the first quadrant and another tangent is drawn to the vertex $A$ of the parabola. Let both the tangent meet at a point $B$,if area of the triangle $ABP=32{\text{unit}}^2$, then equation of the tangent is

• A. $4y=x+8$
• B. $2y=x+8$
• C. $4y=x+16$
• D. $4x=y+16$

Answer 1

The correct option is C $4y=x+16$

Let the coordinates of $P\equiv (t^2,2t)$
For the given parabola tangent at vertex will be $y-$axis
equation of the tangent to the parabola at $P$ will be
$yt=x+t^2$
for the coordinates of $B;x=0$
$\therefore y=t\therefore B\equiv (0,t)$

So the required area will be
$\frac{t^2\cdot t}{2}=32\Rightarrow t^3=2^6\Rightarrow t=4$
So the equation of the tangent will be
$4y=x+16$