Given slope m- find the equation of normal having slope m to the parabola y2-4 axA- y-m x-2 a m-a m3B- y-m x-2 a m-a m3C- y-m x-2 a m-m3D- y-m x-2 a m-a m3

The correct option is B Given value of normal is m. we know at point (h,k) slope of tangent is 2a/k. Therefore slope of normal would be k/2a. We have, k^2 = ...

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Standard XII

Mathematics

Equation of Normal When Slope Is Given

Given slope m...

Question

Given slope m, find the equation of normal having slope m to the parabola $y^{2} = 4 a x$

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Solution

The correct option is B

Given value of normal is m. we know at point (h,k) slope of tangent is $\frac{2 a}{k}$ . Therefore slope of normal would be $\frac{- k}{2 a}$ .

We have, $k^{2} = 4 a h a n d m = \frac{- k}{2 a}$

Substituting for c in y = mx + c at point (h,k)

k = m(h) + c

$- 2 a (m) = m (\frac{k^{2}}{4 a}) + c$

$- 2 a m - m (\frac{(- 2 a m)^{2}}{4 a}) = c$

we get $c = - 2 a m - a m^{3}$

Therefore equation of normal with slope m for the given parabola would be $y = m x - 2 a m - a m^{3}$

Suggest Corrections

Similar questions

Q. The normal

y = m x - 2 a m - a m^{3}

to the parabola

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subtends a right angle at the vertex if

If the equation of the normal is y = mx + c to the parabola $y^{2} = 4 a x$ , then find the value of 'c' in terms of a and m.

What's the equation of tangent to the parabola $y^{2} = 4 a x$ having a slope 'm'.

Q. What is the equation of the normal with slope m to the ellipse

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is given by $y = m x \pm \frac{m (a^{2} + b^{2})}{\sqrt{a^{2} - b^{2} m^{2}}}$

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Given slope m, find the equation of normal having slope m to the parabola y2=4ax

Given slope m, find the equation of normal having slope m to the parabola $y^{2} = 4 a x$