JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

The equation of the tangent to the curve

y=\sqrt{9-2 x^2}

at the point where the ordinate & the abscissa are equal is

View answer and explanation

Correct option: D

2 x+y-3 \sqrt{3}=0

Option D (

2 x+y-3 \sqrt{{3}}=0

) is correct because it is the equation of the tangent calculated at the point where the ordinate and abscissa are equal on the curve.

More Questions from TANGENTS & NORMALS

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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