The lines tangent to the curves y3- x2y +5y -2x = 0 and x4 - x3y2 + 5x + 2y = 0 at the origin intersect at an angle θ equal to

Question

The lines tangent to the curves y3- x2y +5y -2x = 0 and x4 - x3y2 + 5x + 2y = 0 at the origin intersect at an angle θ equal to (A) (π/6) (B) (π/4) (C) (π/3) (D) (π/2)

Tardigrade Admin · Accepted Answer

Differentiating y3 - x2y + 5y - 2x = 0 w.r.t. x, we get 3y2y' - 2xy - x2y' + 5 y' - 2 = 0 or y' = (2xy + 2/3y2 - x2 + 5) or y'(0, 0) = (2/5) Differentiating x4 - x3y2 + 5x + 2y = 0 w.r.t. x, we get 4x3 - 3x2y2 - 2x3yy' + 5 + 2y' = 0 or y' = (3x2y2 - 4x3 - 5/2 - 2x3 y) or y'(0,0) = -(5/2). Thus, both the curves intersect at right angle.

Q. The lines tangent to the curves $y^{3} - x^{2} y + 5 y - 2 x = 0$ and $x^{4} - x^{3} y^{2} + 5 x + 2 y = 0$ at the origin intersect at an angle $θ$ equal to

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

Q. The lines tangent to the curves y3−x2y+5y−2x=0 and x4−x3y2+5x+2y=0 at the origin intersect at an angle θ equal to

6π​

4π​

3π​

2π​

Q. The lines tangent to the curves $y^{3} - x^{2} y + 5 y - 2 x = 0$ and $x^{4} - x^{3} y^{2} + 5 x + 2 y = 0$ at the origin intersect at an angle $θ$ equal to

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$