Equation of the tangent to $y^{2} = 6 x$ at the positive end of the latusrectum is

2x-2y+3=0

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2x+2y+3=0

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y=4

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y+4=0

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Solution

The correct option is A
2x-2y+3=0

$y^{2} = 6 x a = \frac{6}{4} = \frac{3}{2}$
$L = (\frac{3}{2}, 3)$
Eqn of tgt is $(y - 3) = (x - \frac{3}{2})$
$\Rightarrow x - y + \frac{3}{2} = 0 \Rightarrow 2 x - 2 y + 3 = 0$

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