Find the equation of tangents drawn to $y^{2} + 12 x = 0$ from the point $(3, 8) .$

3 x - y - 1 = 0

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x + 3 y - 27 = 0.

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x - 3 y - 27 = 0.

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3 x + y - 1 = 0

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Solution

The correct options are
A $3 x - y - 1 = 0$
B $x + 3 y - 27 = 0.$
$y^{2} + 12 x = 0 ∴ y^{2} = - 12 x$
or $4 a = - 12 ∴ a = - 3.$
Any tangent to the parabola $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$ or $y = m x - \frac{3}{m}$
$∵ a = - 3$ and m is unknown
Since the tangent passes through the point $(3, 8),$ we have
$8 = 3 m - \frac{3}{m}$ or $3 m^{2} - 8 m - 3 = 0$
or $(m - 3) (3 m + 1) = 0 ∴ m = 3, - \frac{1}{3} .$
Above gives two values of m and hence there will be two tangents through
the point $(3, 8) .$ On putting the values of m the two tangents are
$3 x - y - 1 = 0$ and $x + 3 y - 27 = 0.$
They are perpendicular.
Ans: A,B

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