JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

The point of the curve

{y^2} = 2\left( {x - 3} \right)

at which the normal is parallel to line

\;y - 2x + 1 = 0

View answer and explanation

Correct option: C

\left( {5,\; - 2} \right)

Option C ((5, -2)) is correct because it is found by equating the slope of the normal to the slope of the given line.

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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