The foci of a hyperbola are (2 , 0 ) and its eccentricity is 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 an

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Solution

Q.
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 _________ . [2023]

Ans.
(12)

$e = \frac{3}{2}, a e = 2$

$\Rightarrow a = \frac{4}{3} \Rightarrow e^{2} = 1 + \frac{b^{2}}{a^{2}}$

$\Rightarrow \frac{5}{4} = \frac{9 b^{2}}{16} \Rightarrow b^{2} = \frac{5}{4} \times \frac{16}{9} = \frac{20}{9}$

$Equation of tangent:$ $y = m x + \sqrt{a^{2} m^{2} - b^{2}}$

$y = \frac{3}{2} x + \sqrt{\frac{16}{9} \cdot \frac{9}{4} - \frac{20}{9}}$

$y = \frac{3}{2} x + \frac{4}{3}$

$y-intercept = \frac{4}{3},$ $x-intercept = \frac{- 8}{9}$

$| 6 a |$ $+$ $| 5 b |$ $=$ $| 6 \times \frac{- 8}{9} |$ $+$ $| 5 \times \frac{4}{3} |$ $= \frac{16}{3} + \frac{20}{3} = \frac{36}{3} = 12$

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