QJEEAdv12Ma0ba3: JEE Advanced 2012 Mathematics - Class 11 JEEAdv12Ma0ba3 Solution: The locus of the mid-point of the chord of contact of tangents…

Equation of the chord AB is $\mathrm{T}=0$

$\Rightarrow \alpha x+\left(\frac{4 \alpha-20}{5}\right) y=9\quad \text{... (i)}$

Also, equation of the chord AB whose mid-point is $(h, k)$ is $\mathrm{T}=\mathrm{S}_1$

$\begin{aligned}& \Rightarrow \quad h x+k y-9=h^2+k^2-9 \\& \Rightarrow \quad h x+k y=h^2+k^2 \quad \text{... (ii)}\end{aligned}$

$\because$ Equations (i) and (ii) both represent the same line

$\begin{aligned}& \therefore \frac{\alpha}{h}=\frac{\frac{4 \alpha-20}{5}}{k}=\frac{9}{h^2+k^2} \\& \Rightarrow \alpha=\frac{9 h}{h^2+k^2} \quad \text { and } \quad \frac{4 \alpha-20}{5}=\frac{9 k}{h^2+k^2}\end{aligned}$

$\begin{aligned}& \Rightarrow \alpha=\frac{9 h}{h^2+k^2} \quad \text { and } \quad 4 \alpha=\frac{45 k}{h^2+k^2}+20 \\& \Rightarrow \alpha=\frac{9 h}{h^2+k^2} \quad \text { and } \quad \alpha=\frac{45 k+20\left(h^2+k^2\right)}{4\left(h^2+k^2\right)}\end{aligned}$

$\begin{array}{ll}\Rightarrow & \frac{9 h}{h^2+k^2}=\frac{45 k+20\left(h^2+k^2\right)}{4\left(h^2+k^2\right)} \\\Rightarrow & 36 h=45 k+20\left(h^2+k^2\right)\end{array}$

$\Rightarrow 20\left(x^2+y^2\right)-36 x+45 y=0$ , which is the required locus.