Find the equation of normal to the parabola $y^{2} = 4 a x$ at point (h,k) on the parabola

$y - k = \frac{- k}{a (x - h)}$

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$y - k = \frac{- k}{2 a (x - h)}$

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$y - k = \frac{k}{2 a (x - h)}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$y - k = \frac{- k}{2 (x - h)}$

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Solution

The correct option is B
$y - k = \frac{- k}{2 a (x - h)}$

We know that the slope of the tangent of the given parabola at (h,k) is $\frac{2 a}{k}$

. ðSlope of normal = $\frac{- 1}{(\frac{2 a}{k})} = \frac{- k}{2 a}$

y - k = $\frac{- k}{2 a (x - h)}$

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