Let a line $L_{1}$ be tangent to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ and let $L_{2}$

Question

Let a line $L_{1}$ be tangent to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ and let $L_{2}$ be the line passing through the origin and perpendicular to $L_{1}$. If the locus of the point of intersection of $L_{1}$ and $L_{2}$ is $\left(x^{2}+y^{2}\right)=\alpha x^{2}+\beta y^{2}$, then $\alpha+\beta$ is equal to

Professor PC Thomas Classes · Answer

(12) : Given hyperbola is $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ Equation of tangent line $L_{1}$ is\ $y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}=m x \pm \sqrt{16 m^{2}-4}$\ Equation of line $\perp^{r}$ to line $L_{1}$ is $y=-\frac{x}{m}$\ Put $m=\frac{-x}{y}$ in line $L_{1}$;\ $y=-\frac{x^{2}}{y} \pm \sqrt{16\left(\frac{x^{2}}{y^{2}}ight)-4}$\ $y+\frac{x^{2}}{y}= \pm \sqrt{\frac{16 x^{2}-4 y^{2}}{y^{2}}}$\ $\Rightarrow\left(\frac{y^{2}+x^{2}}{y}ight)^{2}=\frac{16 x^{2}-4 y^{2}}{y^{2}}$\ $\left(y^{2}+x^{2}ight)^{2}=16 x^{2}-4 y^{2}$\ Comparing, $\alpha=16, \beta=-4$. Then, $\alpha+\beta$ $=16-4=12$\