The point $ (0,5) $ is closest to the curve $ x^{2}=2 y $ at

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

The point

(0,5)

is closest to the curve

x^{2}=2 y

View answer and explanation

Correct option: D

(2 \sqrt{2}, 4)

Option D (

(2 \sqrt{{2}}, 4)

) is correct because by minimizing the distance between the point (0,5) and a general point on the curve, we find the closest point.

Practice All Chapter MCQs →

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