Question

Equation of a trangent to the curve $y\cot x=y^3\tan x$ at the point where the abscissa is $\frac{\pi }{4}$ is

• A. $4x+2y=\pi +2$
• B. $4x-2y=\pi +2$
• C. x = 0
• D. y = 0

Answer 1

Answer: (A), (B), (D)

$4x-2y=\pi +2$
$4x+2y=\pi +2$
y = 0

Put $x=\frac{\pi }{4}$ in the given equation,
$y=y^3\Rightarrow y(y^2-1)=0\Rightarrow y=-1,0,1$
So the points are $(\frac{\pi }{4},-1),(\frac{\pi }{4},0)$ and $(\frac{\pi }{4},1)$
Now given equation may be written as, $y={\tan }^{2}x.y^3$
Differentiating w.r.t $x$
$\frac{dy}{dx}={\tan }^{2}x.3y^2\frac{dy}{dx}+y^3.2\tan x.{\sec }^{2}x$
$\Rightarrow \frac{dy}{dx}=\frac{2y^3{\sec }^{2}x.\tan x}{1-3{\tan }^{2}x.y^2}$
Thus slope of tangents are,
$m_1={\left(\frac{dy}{dx}\right)}_{(\frac{\pi }{4},-1)}=\frac{2(-1).2.1}{1-\mathrm{3.1.1}}=2$
$m_2={\left(\frac{dy}{dx}\right)}_{(\frac{\pi }{4},0)}=\frac{2\left(0\right).2.1}{1-\mathrm{3.1.0}}=0$
$m_3={\left(\frac{dy}{dx}\right)}_{(\frac{\pi }{4},1)}=\frac{2\left(1\right).2.1}{1-\mathrm{3.1.1}}=-2$
Hence equation of tangents are,
$(y+1)=2(x-\frac{\pi }{4})\Rightarrow 4x-2y=\pi +2$
$(y-0)=0(x-\frac{\pi }{4})\Rightarrow y=0$
$(y-1)=2(x-\frac{\pi }{4})\Rightarrow 4x+2y=\pi +2$