Hyperbola

Let $(h, k)$ be the mid-point of a chord of the circle $x^2+y^2=16$.

Then, the equation of the chord is

$ \begin{array}{rlrl} & h x+k y-16 =h^2+k^2-16 \\& h x+k y =h^2+k^2 \\ \Rightarrow & y =\left(\frac{-h}{k}\right)+\left(\frac{h^2+k^2}{k}\right) \end{array} $

This will be a tangent to the hyperbola $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$, if

$ \begin{aligned} & \left(\frac{h^2+k^2}{k}\right)^2=4^2 \times\left(\frac{-h}{k}\right)^2-3^2\left[\text { using } c^2=a^2 m^2-b^2\right] \\ & \Rightarrow\left(h^2+k^2\right)^2=16 h^2-9 k^2 \end{aligned} $

Hence, the locus of $(h, k)$ is

$ \left(x^2+y^2\right)^2=16 x^2-9 y^2=k x^2-l y^2 $

Thus, $k=16, l=9$ and $k+l=16+9=25$

Given, Hyperbola $ =4 x^{2}-9 y^{2}-1=0 $

$ \Rightarrow \quad \frac{x^{2}}{\left(\frac{1}{4}\right)}-\frac{y^{2}}{\left(\frac{1}{9}\right)}=1 $

$ \Rightarrow \quad \frac{x^{2}}{\left(\frac{1}{2}\right)^{2}}-\frac{y^{2}}{\left(\frac{1}{3}\right)^{2}}=1 $

We know that, if hyperbola is $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $, then

Eccentricity, $ e=\sqrt{1+\frac{b^{2}}{a^{2}}} $

$ =\sqrt{1+\frac{\frac{1}{9}}{\frac{1}{4}}}=\frac{\sqrt{13}}{3} $

and focus $ \equiv( \pm a e, 0)=\left( \pm \frac{1}{2} \times \frac{\sqrt{13}}{3}, 0\right)=\left( \pm \frac{\sqrt{13}}{6}, 0\right) $

x = 9 meets the hyperbola at (9, 6$\sqrt2$) and (9, $-$6$\sqrt2$). Then, the equations of tangent at these points are 3x $-$ 2$\sqrt{2y}$ $-$ 3 = 0. The combined equation of these two tangent is 9x² $-$ 8y² $-$ 18x + 9 = 0