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