Question

A particle moves along the curve $y=x^2+2x$. Then the points on the curve are the x and y coordinates of the particle changing at the same rate, are.

• A. $(\frac{-3}{4},\frac{-1}{2})$
• B. $(\frac{-1}{2},\frac{-3}{4})$
• C. $(\frac{3}{4},\frac{1}{2})$
• D. $(\frac{1}{2},\frac{3}{4})$

Answer 1

The correct option is B $(\frac{-1}{2},\frac{-3}{4})$

$y=x^2+2x$
given $\frac{dx}{dt}=\frac{dy}{dt}..\left(1\right)$
Now $y=x^2+2x$
$\Rightarrow \frac{dy}{dt}=(2x+2)\times \frac{dx}{dt}$
$\Rightarrow 2x+2=1$ using (1)
$\therefore x=-\frac{1}{2};y=-\frac{3}{4}$
Hence the points is $(\frac{-1}{2},\frac{-3}{4})$