The coordinates of the point of contact of the tangent to the parabola $y^{2} = 16 x$ , which is perpendicular to the line $2 x - y + 5 = 0$ are

(16, 16)

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(16, -16)

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(1, 4)

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(1, -4)

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Solution

The correct option is B
(16, -16)

$y^{2} = 16 x . . . . . . . . . (1) 4 a = 16 \Rightarrow a = 4 2 x - y + 5 = 0 . . . . . . . (2)$
A line is perpendicular to the line (2) is
$x + 2 y + λ = 0 . . . . . . (3) \Rightarrow S l o p e, m = \frac{- 1}{2}$
Slope of the tangent to a parabola at $(a t^{2}, 2 a t)$ is $\frac{1}{t}$ . Hence, a line having slope m intersects the parabola at $(\frac{a}{m^{2}}, \frac{2 a}{m})$
Line (3) touches the parabola (1). So, the coordinates of the point of contact is
$\Rightarrow P (\frac{a}{m^{2}}, \frac{2 a}{m}) = (16, - 16)$

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Similar questions

The coordinates of the point of contact of the tangent to the parabola $y^{2} = 16 x$ , which is perpendicular to the line $2 x - y + 5 = 0$ are

y^{2} = 16 x

2 x - y + 5 = 0.

y^{2} = 16 x

The coordinates of the point of contact of the tangent to the parabola $y^{2} = 16 x$ , which is perpendicular to the line $2 x - y + 5 = 0$ are

y^{2} = 16 x

2 x - y + 5 = 0

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