Show that the curves $6 x^{2} - 5 x + 2 y = 0$ and $4 x^{2} + 8 y^{2} = 3$ touch each other at $(\frac{1}{2}, \frac{1}{2})$

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Detailed Solution

Given curves

$6 x^{2} - 5 x + 2 y = 0 . . . . . . . (1)$

$4 x^{2} + 8 y^{2} = 3 . . . . . . . (2)$

Diff (1) and (2) w.r. to ‘ $x$ ’

$12 x - 5 + 2 \frac{d y}{d x} = 0 8 x + 16 y \frac{d y}{d x} = 0$

$2 \frac{d y}{d x} = 5 - 12 x 16 y \frac{d y}{d x} = - 8 x$

$\frac{d y}{d x} = \frac{5 - 12 x}{2} \frac{d y}{d x} = \frac{- 8 x}{16 y} = \frac{- x}{2 y}$

$m_{1} = {(\frac{d y}{d x})}_{(\frac{1}{2}, \frac{1}{2})} = \frac{5 - 12 (\frac{1}{2})}{2} = \frac{5 - 6}{2} = \frac{- 1}{2}$

$m_{2} = {(\frac{d y}{d x})}_{(\frac{1}{2}, \frac{1}{2})} = \frac{- 1 / 2}{2 (\frac{1}{2})} = \frac{- 1}{2}$

Here $m_{1} = m_{2}$

The given curves touch each other at $(\frac{1}{2}, \frac{1}{2})$

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