Equation of a common tangent with positive slope to the circle as well as to the hyperbola is [2010]

Question

The circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points $A$ and $B$ . [2010]

Q. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Smart Achievers · Accepted Answer

(2)

Any tangent to $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ is $\frac{x \sec α}{3} - \frac{y \tan α}{2} = 1$

It touches circle with center $(4, 0)$ and radius $= 4$

$∴$ $\frac{\frac{4 \sec α - 3}{3}}{\sqrt{\frac{\sec^{2} α}{9} + \frac{\tan^{2} α}{4}}} = 4$

$\Rightarrow 16 \sec^{2} α - 24 \sec α + 9 = 144 (\frac{\sec^{2} α}{9} + \frac{\tan^{2} α}{4})$

$\Rightarrow 12 \sec^{2} α + 8 \sec α - 15 = 0$ $\Rightarrow \sec α = \frac{5}{6} or - \frac{3}{2}$

since $\sec α = \frac{5}{6} < 1$ is not possible.

$∴$ $\sec α = - \frac{3}{2}$ $\Rightarrow \tan α = \pm \frac{\sqrt{5}}{2}$

$∴$ $Slope of tangent = \frac{2 \sec α}{3 \tan α} = \frac{2 (- 3 / 2)}{3 (- \sqrt{5} / 2)} = \frac{2}{\sqrt{5}}$ $(for +ve value of \tan α)$

$∴$ $Equation of tangent is$ $\frac{- x}{2} + \frac{y \sqrt{5}}{4} = 1$

$\Rightarrow 2 x - \sqrt{5} y + 4 = 0$

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