Question

Find the points on the curve $y=x/(1-x^2)$ where the tangents makes an angle of $\pi /4$ with x-axis

• A. $(\sqrt{3},-\sqrt{\frac{2}{3}}),(-\sqrt{2},\sqrt{\frac{2}{3}})$
• B. $(\sqrt{3},-\sqrt{\frac{3}{4}}),(-\sqrt{3},\sqrt{\frac{3}{4}})$
• C. $(\sqrt{3},-\sqrt{\frac{3}{2}}),(-\sqrt{3},\sqrt{\frac{3}{2}})$
• D. none of these

Answer 1

Answer: (C)

$(\sqrt{3},-\sqrt{\frac{3}{2}}),(-\sqrt{3},\sqrt{\frac{3}{2}})$

$\frac{dy}{dx}=tan{45}^0$
Or
$\frac{dy}{dx}=1$
Or
$\frac{(1-x^2)-x(-2x)}{(1-x^2{)}^2}=1$
Or
$1-x^2+2x^2=(1-x^2{)}^2$
Or
$1+x^2=1-2x^2+x^4$
Or
$x^4-3x^2=0$
Or
$x^2[x^2-3]=0$
$x=0$ or
$x=\pm \sqrt{3}$
Hence
$y=\frac{\sqrt{3}}{1-3}=-\frac{\sqrt{3}}{2}$
$y=\frac{-\sqrt{3}}{1-3}=\frac{\sqrt{3}}{2}$
Hence
$(\sqrt{3},-\frac{\sqrt{3}}{2}),(-\sqrt{3},\frac{\sqrt{3}}{2})$.