JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 8)
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 :
($$-$$2, 8)
(2, 8)
(1, 5)
(3, 13)
Explanation
Given, curve y = x2 + 4
and, line y = 4x $$-$$ 1
Here, y = x2 + 4
_24th_February_Evening_Shift_en_8_1.png)
$$\therefore$$ $${{dy} \over {dx}} = 2x$$ ..... (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)
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)
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