Question

The two tangents to the curve $a{x}^2+2hxy+b{y}^2=1,a>0$ at the points where it crosses the X-axis, are

• A. parallel
• B. perpendicular
• C. inclined at an angle $\frac{\pi }{4}$
• D. None of these

Answer 1

The correct option is A parallel

$a{x}^2+2hxy+b{y}^2=1$

At $y=0$

$a{x}^2+0+0=1\Rightarrow x=\pm \frac{1}{\sqrt{a}}\frac{d}{dx}(a{x}^2+2hxy+b{y}^2=1)2ax+2hy+2hx\frac{dy}{dx}+2by\frac{dy}{dx}=0$

At $(\pm \frac{1}{\sqrt{a}},0)$

$2ax+0+2hx\frac{dy}{dx}+0=02a(\pm \frac{1}{\sqrt{a}})+2h(\pm \frac{1}{\sqrt{a}})\frac{dy}{dx}=0a+h\frac{dy}{dx}=0\frac{dy}{dx}=-\frac{a}{h}$

Slope is same , so the tangents are parallel

So option$A$ is correct