JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Analyzing +4 | -1

If the curves

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

and

y^3=8 x

intersect at right angles, then the value of

a^2

is equal to

View answer and explanation

Correct option: D

Option D (4) is correct because it is the value of

a^2

that satisfies the condition for the curves to intersect at right angles.

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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