Question

Through the vertex $O$ of the parabola $y^2=4ax$ chords $OP,OQ$ are drawn at right angles to each other. If the locus of mid points of the chord $PQ$ is a parabola, then its vertex is

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

Answer 1

Answer: (A)

$(4a,0)$

Since the locus of the mid point of the chord of PQ is the parabola, hence its vertex is the point where the chord PQ meets the axis of the parabola.

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

Since the vertex of the original (given) parabola is $(0,0)$, therefore the equation of OP is

$\frac{y-0}{x-0}=\frac{y-2at}{x-a{t}^2}$

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

Since OQ is perpendicular to OP and it two passes through the origin, therefore, equation of OQ 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 $OQ$ 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 vertex of the new parabola 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 vertex is $(4a,0)$.