Question

The length of perpendicular from (0, 0) to the tangent drawn

to the curve $y^{2} = 4(x + 2)$ at point (2, 4) is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{3}{\sqrt{5}}$
• C. $\frac{6}{\sqrt{5}}$
• D. 1

Answer 1

Answer: (C)

$\frac{6}{\sqrt{5}}$

Answer 2

Differentiating the given equation w.r.t. x , $2y\frac{dy}{dx} = 4$ at point

(2, 4) $\frac{dy}{dx} = \frac{1}{2}$

$P = \frac{y_{1} - x_{1}\left( \frac{dy}{dx} \right)}{\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}}$ = $\frac{4 - 2\left( \frac{1}{2} \right)}{\sqrt{1 + \frac{1}{4}}} = \frac{6}{\sqrt{5}}$.

$f^{'}(x) \geq 0,$ if $x \in \left\lbrack - \frac{1}{2},1 \right\rbrack$, because $e^{x(1 - x)}$ is always positive. So, $f(x)$ is increasing on $\left\lbrack - \frac{1}{2},1 \right\rbrack$.