Question

For the curve $x^2+4xy+8y^2=64$ the tangents are parallel to the x-axis only 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. $(8,0)$ and $(-8,0)$

Answer 1

Answer: (B)

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

The given curve is $x^2+4xy+8y^2=64$ ....$\left(1\right)$

If the tangents are parallel to $x$ - axis then the slope of the curve at those points is zero.

For the slope of the curve differentiating the given equation of curve w.r.t. $x$, we get,

$\Rightarrow 2x+4y+4x\frac{dy}{dx}+8\left(2y\right)\frac{dy}{dx}=0$ ....$\left(2\right)$

As tangents are parallel to $x$ - axis, so $\frac{dy}{dx}=0$

Putting the value of the slope of the curve in equation $\left(2\right)$, we get,

$\Rightarrow x+2y=0$ or $x=-2y$

Now putting the value of $x$ into equation $\left(1\right)$, we get,

$\Rightarrow 4y^2-8y^2+8y^2=64$

$\Rightarrow y=\pm 4$

As we know that $x=-2y$ from previous calculations so $x=-2(\pm 4)$

$\Rightarrow x=\mp 8$

Hence tangents to the curve are parallel to $x$ - axis only at $(\mp 8,\pm 4)$ or at $(8,-4)$ and $(-8,4)$

Hence correct option is $B$.