Equation of tangent to the parabola $y^{2} = 16 x$ at $P (3, 6)$ is

4 x - 3 y + 12 = 0

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3 y - 4 x - 12 = 0

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4 x - 3 y - 24 = 0

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3 y - x - 24 = 0

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Solution

The correct option is B $3 y - 4 x - 12 = 0$
Equation of tangent to the parabola $y^{2} = 4 a x$ at point $(x_{1}, y_{1})$ is given as
$y y_{1} = 2 a (x + x_{1}) \Rightarrow 6 y = 8 (x + 3) \Rightarrow 3 y - 4 x - 12 = 0$

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The line 4x - 3y + 12 = 0 meets x-axis at A. Write the co-ordinates of A.

Determine the equation of the line through A and perpendicular to 4x - 3y + 12 = 0.

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