The point (0, 3) is nearest to the curve ${x^2} = 2y$ at

$\left( {2\sqrt 2 ,\;0} \right)$

JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

1 Medium

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Application +4 | -1

The point (0, 3) is nearest to the curve

{x^2} = 2y

View answer and explanation

Correct option: C

(2, 2)

Option C (2, 2) is correct because it represents the point on the curve

{x^2} = 2y

that minimizes the distance to the point (0, 3).

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\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3