The foci of hyperbola are( $\pm$ 2, 0) and its eccentricity is 3/2. A tangent, perpendicular to the line 2x + 3y = 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______

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answer is 12.

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Detailed Solution

$\begin{array}{l} S & S^{'} = (\pm 2, 0), e = \frac{3}{2} \\ 2 ae = 4 \Rightarrow ae = 2 \Rightarrow a \cdot \frac{3}{2} = 2 \Rightarrow a = \frac{4}{3} \\ b^{2} = a^{2} e^{2} - a^{2} \Rightarrow b^{2} = 4 - \frac{16}{9} = \frac{20}{9} \end{array}$
Tangent is 3x - 2y + k = 0
$\begin{array}{l} y = \frac{3 x}{2} + \frac{k}{2} \\ c^{2} = a^{2} m^{2} - b^{2} \Rightarrow \frac{16}{9} \times \frac{9}{4} - \frac{20}{9} = \frac{k^{2}}{4} \\ \Rightarrow \frac{144 - 80}{9 \times 4} = \frac{k^{2}}{4} \Rightarrow k = \pm \frac{8}{3} \\ 3 x - 2 y \pm \frac{8}{3} = 0 \\ 3 x - 2 y = \frac{8}{3} \Rightarrow \frac{x}{8 / 9} = \frac{y}{- 8 / 6} = 1 \\ | 6 a | + | 5 b | = 6 \cdot \frac{8}{9} + 5 \cdot \frac{8}{6} = 12 \end{array}$