Question

If $x^2+2xy+2y^2=1,$ then $\frac{dy}{dx}$ at the point where $y=1$ is equal to

• A. 1
• B. 2
• C. -1
• D. -2
• E. 0

Answer 1

Answer: (E)

Given, $x^2+2xy+2y^2=1\cdots$
Put $y=1$,
$x^2+2x(1)+2(1{)}^2=1$
$\Rightarrow x^2+2x+1=0$
$\Rightarrow (x+1{)}^{2}0$
$\Rightarrow x=-1$
On differentiating Eq. (i) w.r.t. $x$, we get
$2x+2x\frac{dy}{dx}+2y+4y\frac{dy}{dx}=0$
$\Rightarrow 2\frac{dy}{dx}(x+2y)=-2(y+x)$
$\Rightarrow \frac{dy}{dx}=\frac{-(y+x)}{x+2y}$
When $x=-1,y=1$
$\frac{dy}{dx}=-\frac{(1-1)}{-1+2}=0$