The equation of the two tangents drawn from $(1, 4)$ to the parabola $y^{2} = 12 x$ are

x - y + 3 = 0, 3 x - y + 1 = 0

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x - y + 1 = 0, x - 2 y + 4 = 0

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x + y - 2 = 0, x - y = 3

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x + y - 1 = 0, x + 2 y + 4 = 0

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Solution

The correct option is B $x - y + 3 = 0, 3 x - y + 1 = 0$
Equation of tangent to the parabola $y^{2} = 12 x$ is given by,
$y = m x + \frac{3}{m}$
Given it passes through $(1, 4)$
$\Rightarrow 4 = m + \frac{3}{m} \Rightarrow m^{2} - 4 m + 3 = 0$
$\Rightarrow (m - 1) (m - 3) = 0 \Rightarrow m = 1, 3$
Hence, required tangents are,
$y = x + 3$ and $y = 3 x + 1$

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Similar questions

(1, 4)

y^{2} = 12 x

x^{2} + y^{2} - 2 x + 4 y = 0

(4, 10)

y^{2} = 9 x

x^{2} = 2 y

y^{2} = 4 x

x^{2} = 4 y

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