If P is a point on the parabola y = x2 + 4 which is closest to the straight line y = 4x - …

Question

If P is a point on the parabola y = x2 + 4 which is closest to the straight line y = 4x - 1, then the co-ordinates of P are :

Accepted Answer

Given, curve y = x2 + 4

and, line y = 4x - 1

Here, y = x2 + 4

$$\therefore$$ $${{dy} \over {dx}} = 2x$$ ..... (i)

and y = 4x - 1

$${{dy} \over {dx}} = 4$$ ..... (ii)

Let the required point be P(x1, y1).

$$\therefore$$ $${\left. {{{dy} \over {dx}}} \right|_P} = 2{x_1}$$ ..... (iii)

$$\because$$ Slopes will be equal.

$$\therefore$$ 2x1 = 4 [from Eqs. (ii) and (iii)]

$$ \Rightarrow {x_1} = {4 \over 2} = 2$$

Now, the given point P(x1, y1) lies on curve y = x2 + 4,

$$\therefore$$ y1 = x$_1^2$ + 4

$$\Rightarrow$$ y1 = 22 + 4 = 8

Hence, required coordinate of P = (2, 8)

Solutions · 1