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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{π}{2}$
Given curve is
$y^{3} - x^{2} y + 5 y - 2 x = 0$
Slope of tangent to curve at $(0, 0) = m_{1} = \frac{2}{5}$
Given curve II as
$x^{4} - x^{3} y^{2} + 5 x + 2 y = 0$
Slope of tangent to curve at $(0, 0) = m_{2} = - \frac{5}{2}$
Since $m_{1} . m_{2} = - 1$
Hence , angle between tangent to the curves is $\frac{π}{2}$

Suggest Corrections

Similar questions

y^{3} - x^{2} y + 5 y - 2 x = 0

x^{4} - x^{3} y^{2} + 5 x + 2 y = 0

θ

x^{3} - y^{3} + x^{2} y - y x^{2} + 3 x - 2 y = 0

x^{5} - y^{4} + 2 x + 3 y = 0

θ

The straight lines joining the origin to the points of intersection of the line 2x + y = 1 and curve $3 x^{2} + 4 x y - 4 x + 1 = 0$ include an angle

y = x^{2} - 5 x + 6

x^{3} - 3 x y^{2} + 2 = 0

3 x^{2} y - y^{3} = 2

Join BYJU'S Learning Program