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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 $..........$.
$11$
$12$
$13$
$10$
Solution
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$ae =2$ and $e =\frac{3}{2} \Rightarrow a =\frac{4}{3}$
also $b ^2= a ^2 e ^2- a ^2 \Rightarrow 4-\frac{16}{9}$
$\Rightarrow b^2=\frac{20}{9}$
$\text { Slope of tangent }=\frac{3}{2}$
So tangent equation will be
$y = mx \pm \sqrt{a^2 m^2-b^2}$
$\Rightarrow y =\frac{3 x}{2} \pm \sqrt{\frac{16}{9} \cdot \frac{9}{4}-\frac{20}{9}}$
$\Rightarrow y =\frac{3 x}{2} \pm \frac{4}{3} \Rightarrow\left| x _{\text {intercept }}\right|=\frac{8}{9}$
$\left|y_{\text {intercept }}\right|=\frac{4}{3}$
$\Rightarrow|6 a|+|5 b|=\frac{48}{9}+\frac{60}{9}=\frac{109}{9}=12$
