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# The point on the curve x2 = 2y which is nearest to the point (0, 5) is (A)  (B)  (C) (0, 0) (D) (2, 2)

The given curve is x2 = 2y. For each value of x, the position of the point will be Let P and A(0, 5) are the given points.Now distance between the points P and A is given by, Let us denote PA2 by Z. Then, Z = Differentiating both sides with respect to y, we get For maxima or minima, we have Now,...

The given curve is x2 = 2y.

For each value of x, the position of the point will be

Let P $\left&space;(&space;x.&space;\frac{x^{2}}2{}&space;\right&space;)$ and A(0, 5) are the given points.
Now distance between the points P and A is given by,

Let us denote PA2 by Z. Then,
Z = $y^{2}$$-&space;8y&space;+25$

Differentiating both sides with respect to y, we get

$\frac{dZ}{dy}=2y-8$

For maxima or minima, we have

$\frac{dZ}{dy}=&space;0$

$\Rightarrow&space;2y-8=0$

$\Rightarrow&space;y=4$

$\left&space;[&space;\frac{d^{2}Z}{dy^{2}}&space;\right&space;]_{y=4}=2>&space;0$

Now, $x^{2}=2y$

⇒ $x^{2}$ = 2×4

$x^{2}=8$

$x=&space;2\sqrt{2}$ or $x=-2\sqrt{2}$

So, Z is minimum at $\left&space;(&space;2\sqrt{2},4&space;\right&space;)$ or $\left&space;(&space;-2\sqrt{2},&space;4\right&space;)$

Or, PA2 is minimum at $\left&space;(&space;2\sqrt{2},4&space;\right&space;)$  or $\left&space;(&space;-2\sqrt{2},&space;4\right&space;)$

Or, PA is minimum at  $\left&space;(&space;2\sqrt{2},4&space;\right&space;)$  or  $\left&space;(&space;-2\sqrt{2},&space;4\right&space;)$

So, distance between the points and A(0, 5is minimum at $\left&space;(&space;2\sqrt{2},4&space;\right&space;)$ or $\left&space;(&space;-2\sqrt{2},&space;4\right&space;)$

So, the correct answer is A.

(A) (2√2, 4)

I am expert in electrical engineering . I have b tech and m tech from nit warangal total experi 10 y

A

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