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Q.
The lines tangent to the curves $y^{3}- x^{2}y +5y -2x = 0$ and $x^{4} - x^{3}y^{2} + 5x + 2y = 0$ at the origin intersect at an angle $\theta$ equal to
Application of Derivatives
Solution:
Differentiating $y^3 - x^2y + 5y - 2x = 0$ w.r.t. $x$, we get
$3y^2y' - 2xy - x^2y' + 5 y' - 2 = 0$
or $y' = \frac{2xy + 2}{3y^2 - x^2 + 5}$ or $y'_{(0, 0)} = \frac{2}{5}$
Differentiating $x^4 - x^3y^2 + 5x + 2y = 0$ w.r.t. $x$, we get
$4x^3 - 3x^2y^2 - 2x^3yy' + 5 + 2y' = 0$
or $y' = \frac{3x^2y^2 - 4x^3 - 5}{2 - 2x^3 y}$ or $y'_{(0,0)} = -\frac{5}{2}$.
Thus, both the curves intersect at right angle.