The foci of a hyperbola are $( \pm 2,0)$ and its eccentricity is $\frac{3}{2}$. A tangent perpendicu

Question

The foci of a hyperbola are $( \pm 2,0)$ and its eccentricity is $\frac{3}{2}$. A tangent perpendicular to the line $2 x+3 y=6$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $x$ - and $y$-axes are $a$ and $b$ respectively, then $|6 a|+|5 b|$ is equal to

Professor PC Thomas Classes · Answer

(12) : $e=\frac{3}{2}, a e=2$\ $\Rightarrow a=\frac{4}{3} \Rightarrow e^{2}=1+\frac{b^{2}}{a^{2}}$\ $\Rightarrow \frac{5}{4}=\frac{9 b^{2}}{16} \Rightarrow b^{2}=\frac{5}{4} 	imes \frac{16}{9}=\frac{20}{9}$\ Equation of tangent :\ $y=m x+\sqrt{a^{2} m^{2}-b^{2}}$\ $y=\frac{3}{2} x+\sqrt{\frac{16}{9} \cdot \frac{9}{4}-\frac{20}{9}}$\ $y=\frac{3}{2} x+\frac{4}{3}$\ $y$-intercept $=\frac{4}{3}, x$-intercept $=\frac{-8}{9}$\ $|6 a|+|5 b|=\left|6 	imes \frac{-8}{9}ight|+\left|5 	imes \frac{4}{3}ight|=\frac{16}{3}+\frac{20}{3}=\frac{36}{3}=12$\