Question

For the curve $x^2+4xy+8y^2=64$ the tangents are parallel to the x-axis only at the points

• A. $\left(0, 2\sqrt{2}\right)$ and $\left(0, -2\sqrt{2}\right)$
• B. $(8,-4)$ and $(-8,4)$
• C. $\left(8\sqrt{2}, -2\sqrt{2}\right)$ and $\left(-8\sqrt{2}, 2\sqrt{2}\right)$
• D. $(8,0)$ and $(-8,0)$

Answer 1

Answer: (B)

$(8,-4)$ and $(-8,4)$

Answer 2

Given curve is, $x^{2}+4 x y+8 y^{2}=64 \quad$...(i)
On differentiating w.r.t $x$, we get
$2 x+4\left(y+x \frac{d y}{d x}\right)+16 y \frac{d y}{d x}=0$
$\Rightarrow 2 x+4 y+(4 x+16 y) \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=-\frac{(x+2 y)}{2(x+4 y)}$
Since, tangent are parallel to $x$-axis only.
i.e., $\frac{d y}{d x}=0$
$\Rightarrow -\frac{(x+2 y)}{2(x+4 y)}=0$
$\Rightarrow x+2 y=0$ ...(ii)
Now, on putting the valus of $x$ from Eqs. (i) in (ii), we get
$4 y^{2}-8 y^{2}+8 y^{2}=64$
$\Rightarrow y^{2}=16$
$\Rightarrow y=\pm 4$
From E (ii) When $y=4, x=-8$
and when $y=-4, x=8$
Hence required points are $(-8,4)$ and $(8-4)$