Question

Through the vertex $A$ of the parabola $y^2=4ax$ chords $AP$ and $AQ$ are drawn at right angles to one another. Then the line $PQ$ meets the axis in a fixed point with coordinates

• A. $(3a,0)$
• B. $(2a,0)$
• C. $(6a,0)$
• D. $(4a,0)$

Answer 1

The correct option is D $(4a,0)$

Let the point P be $(a{t}^2,2at)$

Since the vertex of the above parabola is $(0,0)$, therefore the equation of AP is $\frac{y-0}{x-0}=\frac{y-2at}{x-a{t}^2}$

$y=\frac{2}{t}x$ ...(i)

Since AQ is perpendicular to AP and it two passes through the origin, therefore, equation of AQ is

$y=\frac{-t}{2}x$ ...(ii)

Substituting the $y=\frac{-t}{2}x$ into the equation of the parabola, we can determine the intersection point of $AQ$ and the parabola.

Therefore

$\frac{t^2{x}^2}{4}=4ax$

$x{t}^2=16a$
$x=\frac{16a}{t^2}$

Hence $y=\frac{-8a}{t}$

Therefore the equation of the PQ wil be

$\frac{y+\frac{8a}{t}}{x-\frac{16a}{t^2}}=\frac{y-2at}{x-a{t}^2}$

Since the axis of the parabola is the positive x axis, hence the fixed point will be of the form $(x,0)$

Substituting $y=0$ in the above equation, we get

$\frac{8a}{t}(x-a{t}^2)=(-2at)(x-\frac{16a}{t^2})$

$8ax-8a^2{t}^2=-2a{t}^{2}x+32a^2$

$8x-8a{t}^2=32a-2t^{2}x$

$x(8+2t^2)=32a+8a{t}^2$
$x(8+2t^2)=4a(8+2t^2)$

Therefore $x=4a$

Hence the point is $(4a,0)$.