Question

The lines tangent to the curve $x^3-y^3+x^{2}y-y{x}^2+3x-2y=0$ and $x^5-y^4+2x+3y=0$ at the origin intersect at an angle $\theta$ equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

$\frac{\pi }{2}$

Given equation of curve $x^3-y^3+x^{2}y-y{x}^2+3x-2y=0$
Differentiating we get
$3x^2-3y^2\frac{dy}{dx}+x^2\frac{dy}{dx}+2xy-2xy-x^2\frac{dy}{dx}+3-2\frac{dy}{dx}=0$
$(3y^2+2)\frac{dy}{dx}=3x^2+3$
$\Rightarrow \frac{dy}{dx}=\frac{3x^2+3}{(3y^2+2)}$
Slope of tangent to curve at (0,0) is $m_1=\frac{3}{2}$
Given equation of other curve $x^5-y^4+2x+3y=0$
On differentiation ,
$5x^4-4y^3\frac{dy}{dx}+2+3\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{5x^4+2}{(4y^3-3)}$
Slope of tangent to the curve at (0,0) is $m_2=-\frac{2}{3}$
Here, $m_1{m}_2=-1$
Hence, the lines are perpendicular.