Question

If the tangent at the point $P\left(x_1,y_1\right)$ to the parabola $y^2=4ax$ meets the parabola $y^2=4a(x+b)$ at $Q$ \& $R$, then the mid point of $QR$ is :

• A. $\left(x_1+b,y_1+b\right)$
• B. $\left(x_1-b,y_1-b\right)$
• C. $\left(x_1,y_1\right)$
• D. $\left(x_1+b,y_1\right)$

Answer 1

Answer: (C)

chord of the parabola
$y^2=4a(x+b)\text{ whose middle point is }(h,k)$
$y-k=m(x-h)$.....(1)
$\text{ where }m=\frac{y^{\prime \prime }-y^{\prime }}{x^{\prime \prime }-x^{\prime }}$
Now $\frac{y^{2\prime \prime }-y^{2\prime }}{x^{\prime \prime }-x^{\prime }}=4a$; but $x^{\prime }+x^{\prime \prime }=2h$ and $y^{\prime }+y^{\prime \prime }=2k$
$\frac{y^{\prime \prime }-y^{\prime }}{x^{\prime \prime }-x^{\prime }}=\frac{4a}{y^{\prime \prime }+y^{\prime }}=\frac{4a}{2k}=\frac{2a}{k}$
Hence (1) becomes
$y-k=\frac{2a}{k}(x-h)$
$ky-k^2=2ax-2ah$
$2ax-ky+k^2-2ah=0$.....(2)
Equation of tangent at $P\left(x_1,y_1\right)$ on $y^2=4ax$
$y{y}_1=2a\left(x+x_1\right)$
$2ax-y{y}_1+2a{x}_1=0$.....(3)
comparing (2) and (3)
$1=\frac{k}{y_1}=\frac{k^2-2ah}{2a{x}_1}$
$\Rightarrow k=y_1\text{ and }2a{x}_1=k^2-2ah=y_1^2-2ah=4a{x}_1-2ah\left(ask=y_1\right)$
$\therefore ah=2a{x}_1\Rightarrow h=x_1$
$\text{ Hence }h=x_1\text{ and }k=y_1$