Question

The locus of mid point of the chords of $x^2-y^2=4,$ that also touches the parabola $y^2=8x$ is

• A. $y^2(x-2)=x^3$
• B. $x^2(x-2)=y^3$
• C. $x^3(x-2)=y^2$
• D. $y^3(x-2)=x^2$

Answer 1

The correct option is A $y^2(x-2)=x^3$

Let the mid point of the chord be $P(h,k)$

Equation of the chord to $x^2-y^2=4$ is
$T=S_1\Rightarrow xh-yk=h^2-k^2\Rightarrow y=\frac{hx}{k}+\frac{k^2-h^2}{k}\cdots \left(1\right)$

Equation of tangent to $y^2=8x$ is
$y=mx+\frac{2}{m}\cdots \left(2\right)$
Comparing equation $\left(1\right)$ and $\left(2\right)$, we get
$\frac{h}{k}=m,\frac{k^2-h^2}{k}=\frac{2}{m}\Rightarrow \frac{k^2-h^2}{k}=\frac{2}{\frac{h}{k}}\Rightarrow \frac{k^2-h^2}{k}=\frac{2k}{h}\Rightarrow k^{2}h-h^3=2k^2$

Hence, the required locus is
$y^2(x-2)=x^3$