The equation of the normal at the ends of latus rectum of $y^{2} = 4 a x$ are

x \pm y + 2 a = 0

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x \pm y - 2 a = 0

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x \pm y - 3 a = 0

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x \pm y + 3 a = 0

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Solution

The correct option is A $x \pm y + 2 a = 0$
Coordinates of latus redum $\equiv (2 a, a)$

$\equiv (2 a, - a)$

Now diff eqn : $2 y \frac{d y}{d x} = 4 a$

$∴ \frac{d x}{d y} =$ Slope $= \frac{y}{2 a} = \frac{- a}{2 a} = \frac{- 1}{2}$

Also $\equiv (\frac{- a}{2 a}) = \frac{1}{2}$

$∴$ Equation : $y - a = \frac{- 1}{2} (x - 2 a)$

$∴$ Equation : $x \pm y \pm 2 a = 0$

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y^{2} = 8 x

Find the equations of tangents and normal at the extremities of latus rectum of the parabola $y^{2} = 4 a x .$

y^{2} = 4 b (x - 2 a + b)

x^{2} + 4 a (y - 2 b - a) = 0

x^{2} + 4 a (y - 2 b - a) = 0

y^{2} = 4 b (x - 2 a + b)

y^{2} = 4 a x

P (a, 2 a)

x - y + 1 = 0

a > 0

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