Question

Let 'P' be a point on x² = 4y that is nearest to the point A(0, 4) then co-ordinates of 'P' are

• A. (4, 4)
• B. (0, 0)
• C. (√8, 2)
• D. (2, 1)

Answer 1

Answer: (C)

(√8, 2)

Answer 2

In this case PA must be normal to the given curve. For

x² = 4y, $\frac { d y } { d x } = \frac { x } { 2 }$. Thus equation of normal at P(x₁,y₁) is ;

$\left( y - \frac { x _ { 1 } ^ { 2 } } { 4 } \right) = - \frac { 2 } { x _ { 1 } }$(x – x₁).

It must pass through (0, 4)

⇒ $\left( 4 - \frac { x _ { 1 } ^ { 2 } } { 4 } \right) = \frac { 2 } { x _ { 1 } }$ x₁ = 2

⇒ x₁ =$\pm \sqrt { 8 }$. Apart from this y-axis is also a normal to the curve passing through A(0, 4) and corresponding x₁ = 0.

If x₁ = 0 ⇒ P(0, 0) ⇒ PA = 4

If x₁ = $\pm \sqrt { 8 }$ ⇒ P$( \pm \sqrt { 8 } , 2 )$

⇒ PA = $\sqrt { 8 + 4 } = \sqrt { 12 }$

Thus p ≡ $( \pm \sqrt { 8 } , 2 )$