The point on curve y^2 = 4x which is nearest to the point (2, -8) is Option 1: (3, -3) Option 2: (4, -4) Option 3: (4, 4) Option 4: (9, 6)

Answer: (4, -4). Solution: Parametrise the curve as (t^2, 2t). Then D^2 = (t^2-2)^2 + (2t+8)^2 and d/dtD^2 = 4t^3 - 8t + 8t + 32 = 4t^3 + 32. Setting this to zero gives t = -2, so the nearest point is (4,-4).

Calculus

Application of Derivatives

Medium

core

The point on curve $y^2 = 4x$ which is nearest to the point $(2, -8)$ is

$(3, -3)$

$(4, -4)$

$(4, 4)$

$(9, 6)$

✅ Correct Option: 2

Parametrise the curve as $(t^2, 2t)$ .

Then $D^2 = (t^2-2)^2 + (2t+8)^2$ and $\frac{d}{dt}D^2 = 4t^3 - 8t + 8t + 32 = 4t^3 + 32$ .

Setting this to zero gives $t = -2$ , so the nearest point is $(4,-4)$ .

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(C) Local Maxima of $f(x)$	(III) $-\sqrt{\frac{b}{a}}$
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Choose the correct answer from the options given below:

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The point on curve y2=4xy^2 = 4xy2=4x which is nearest to the point (2,−8)(2, -8)(2,−8) is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The point on curve $y^2 = 4x$ which is nearest to the point $(2, -8)$ is