Question

Let $S$ be the focus of $y^2=4x$ and a point $P$ is moving on the curve such that its abscissa is increasing at the rate of $4$ units/sec, then the rate of increase of projection of $SP$ on $x+y=1$ when $P$ is at $(4,4)$ is

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

Answer 1

The correct option is C $-\sqrt{2}$

$\overrightarrow{V}=(T^2-1)\hat{i}+2T\hat{j}$
Projection of $\overrightarrow{V}$ on $\overrightarrow{n}$
$y=\frac{\overrightarrow{V}\cdot \overrightarrow{n}}{|\overrightarrow{n}|}=\frac{(1-T^2)+2T}{\sqrt{2}}$
Given $\frac{dx}{dt}=4;$ but $x=T^2$
$\Rightarrow \frac{dx}{dt}=2T\frac{dT}{dt}$
When P (4, 4) then $T=2$
$\Rightarrow 4=2\left(2\right)\frac{dT}{dt}$
$\Rightarrow \frac{dT}{dt}=1$
Now $\frac{dy}{dt}=\left(\frac{-2T+2}{\sqrt{2}}\right)\frac{dT}{dt}$
$\therefore$ at $T=2$
$\sqrt{2}\frac{dy}{dt}=-4+2=-2$
$\Rightarrow \frac{dy}{dt}=-\sqrt{2}$