The longest distance of the point (a, 0) from the curve2x2+y2−2x=0 is given by

A
$\sqrt{1 - 2 a + a^{2}}$
B
$\sqrt{1 + 2 a + 2 a^{2}}$
C
$\sqrt{1 + 2 a - a^{2}}$
D
$\sqrt{1 - 2 a + 2 a^{2}}$

Solution:

Let $(x, y)$ be the point on the curve $2 x^{2} + y^{2} - 2 x = 0$

Then its distance from $(a, 0)$ is given by

$\begin{array}{l} S = \sqrt{(x - a)^{2} + y^{2}} \\ \Rightarrow S^{2} = x^{2} - 2 a x + a^{2} + 2 x - 2 x^{2} [Using 2 x^{2} + y^{2} - 2 x = 0] \\ \Rightarrow S^{2} = - x^{2} + 2 x (1 - a) + a^{2} \\ \Rightarrow 2 S \frac{d S}{d x} = - 2 x + 2 (1 - a) \end{array}$

For S to be maximum, we must have

$\frac{d S}{d x} = 0 \Rightarrow - 2 x + 2 (1 - a) = 0 \Rightarrow x = 1 - a$

It can be easily checked that $\frac{d^{2} S}{d x^{2}} < 0 for x = 1 - a$

Hence, Sis maximum for $x = 1 - a$

Putting $x = l - a$ in (i), we get $S = \sqrt{1 - 2 a + 2 a^{2}}$

The longest distance of the point $(a, 0)$ from the curve
$2 x^{2} + y^{2} - 2 x = 0$ is given by

Solution:

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