Question

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

• A. $(0,2\sqrt{2})$ and $(0,-2\sqrt{2})$
• B. $(8,-4)$ and $(-8,4)$
• C. $(8\sqrt{2},-2\sqrt{2})$ and $(-8\sqrt{2},2\sqrt{2})$
• D. $(9,0)$ and $(-8,0)$

Answer 1

The correct option is B $(8,-4)$ and $(-8,4)$

Given curve is $x^2+4xy+8y^2=\mathrm{64.....}\left(i\right)$
On differentiating w.r.t $x$ ,we get
$2x+4(y+x\frac{dy}{dx})+16y\frac{dy}{dx}=0$
$\Rightarrow$ $2x+4y+(4x+16y)\frac{dy}{dx}=0$
$\Rightarrow$ $\frac{dy}{dx}=-\frac{(x+2y)}{2(x+4y)}$
Since, tangent are parallel to $x$-axis only
ie., $\frac{dy}{dx}=0$
$\Rightarrow$ $-\frac{(x+2y)}{2(x+4y)}=0$
$\Rightarrow$ $x+2y=\mathrm{0.....}\left(ii\right)$
Now, on putting the values of $x$ from eqs. $\left(i\right)$ in $\left(ii\right)$ we get
$4y^2-8y^2+8y^2=64$
$\Rightarrow$ $y^2=16$
$\Rightarrow$ $y=\pm 4$
from eq. $\left(ii\right)$
When $y=4,x=-8$
and when $y=-4,x=8$
Hence, required points are $(-8,4)$ and $(8,-4)$