First slide

Hyperbola in conic sections

Question

$A common tangent to the conics x^{2} = 6 y and 2 x^{2} - 4 y^{2} = 9 is$

Moderate

A

$x + y = 1$
B

$x - y = 1$
C

$x + y = \frac{9}{2}$
D

$x - y = \frac{3}{2}$

Solution

$\begin{array}{l} Let y = (m x + c) is tangent to x^{2} = 6 y \\ s o l v e t h e a b o v e e q u a t i o n s \Rightarrow x^{2} = 6 (m x + c) \\ So, x^{2} - 6 m x - 6 c = 0 \\ Put D = b^{2} - 4 a c = 0 \\ \Rightarrow c = \frac{- 3}{2} m^{2} \\ ∴ we get y = m x - \frac{3}{2} m^{2} \dots \dots . (1) \end{array}$
$\begin{array}{l} And given hyperbola equation is 2 x^{2} - 4 y^{2} = 9 \\ \Rightarrow \frac{x^{2}}{\frac{9}{2}} - \frac{y^{2}}{\frac{9}{4}} = 1 \dots . (2) \end{array}$
$Since, equation (1) is a tangent of equation (2) then c^{2} = a^{2} m^{2} - b^{2}$
$\begin{array}{l} \Rightarrow \frac{9}{4} m^{4} = \frac{9}{2} m^{2} - \frac{9}{4} \\ \Rightarrow m^{4} = 2 m^{2} - 1 \\ \Rightarrow m^{4} - 2 m^{2} + 1 = 0 \\ \Rightarrow {(m^{2} - 1)}^{2} = 0 \\ \Rightarrow m = \pm 1 \\ for m = 1, equation of tangent is x - y = \frac{3}{2} \end{array}$
Therefore, the correct answer is (D).

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