# The point on the curve x^{2} = 2y which is nearest to the point (0, 5) is

(A) (2√2, 4) (B) (2√2, 0) (C) (0, 0) (D) (2, 2)

**Solution:**

Maxima and minima are known as the extrema of a function

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

The given curve is x^{2} = 2y.

For each value of x, the position of the point will be (x, x^{2}/2)

Let, (x, x^{2}/2) and A(0, 5) are the given points.

Now distance between the points P and A is given by,

⇒ PA = √(x - 0)² + (x²/2 - 5)²

⇒ PA^{2} = (x - 0)^{2} + (x²/2 - 5)^{2}

⇒ PA^{2} = x^{2} + x^{4}/4 + 25 - 5x^{2}

⇒ PA^{2} = x^{4}/4 - 4x^{2} + 25

⇒ PA^{2} = y^{2} - 8y + 25

(Since x^{2} = 2y)

Let us denote PA^{2 }by Z

Then, Z = y^{2} - 8y + 25

Differentiating both sides with respect to y , we get

dZ/dy = 2y - 8

For maxima or minima, we have

dZ/dy = 0

⇒ 2 y - 8 = 0

⇒ 2 y = 8

⇒ y = 4

d^{2}Z/dy^{2} = 2y

[d^{2}Z/dy^{2}] y = 4 = 2 > 0

⇒ x^{2} = 2y

⇒ x^{2} = 2 x 4

⇒ x^{2} = 8

⇒ x = ± 2√2

So, Z is minimum at (2√2, 4) or (- 2√2, 4).

or, PA^{2} is minimum at (2√2, 4) or (- 2√2, 4).

or, PA is minimum at (2√2, 4) or (- 2√2, 4).

So, distance between the points P (x, x^{2}/2) and A(0, 5) is minimum at (2√2, 4) or (- 2√2, 4).

Thus, the correct option is A

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 27

## The point on the curve x^{2} = 2y which is nearest to the point (0, 5) is (A) (2√2, 4) (B) (2√2, 0) (C) (0, 0) (D) (2, 2)

**Summary:**

The point on the curve x^{2} = 2y which is nearest to the point (0, 5) is (2√2, 4). Thus, the correct option is A

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