JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

If the curves

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{12}} = 1\;

and

{y^3} = 8x\;

intersect at right angle then the value of

{a^2}

is equal to

View answer and explanation

Correct option: D

Option D (4) is correct because the value of

{a^2}

is 4 when the given curves intersect at right angles, which is found by equating the product of their slopes to -1.

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a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z