The angle between the curves $x^{3} + y^{3} + x + 2 y = 0$ and $x y + 2 x = y$ at the origin is

\frac{π}{4}

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

\frac{π}{6}

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 B $\frac{π}{2}$
$3 x^{2} + 3 y^{2} y^{'} + 1 + 2 y^{'} = 0$
$y^{'} = - \frac{(3 x^{2} + 1)}{(3 y^{2} + 2)}$
$y^{'} |_{0, 0} = \frac{1}{2} = m_{1}$
$y + x y^{'} + 2 = y^{1}$
$y^{'} (x - 1) = - (2 + y)$
$y^{'} = (\frac{2 + y}{1 - x})$
$y^{'} |_{(0, 0)} = 2 = m_{2}$
$m_{1} m_{2} = - 1$
$∴$ angle b/w the curves = $\frac{π}{2}$

Suggest Corrections

Similar questions

The angle between the two curves $x^{3} - 3 x y^{2} = 4 and 3 x^{2} y - y^{3} = 4$ is

y^{2} = \frac{2 x}{π}

y = sin x

x = \frac{π}{2}

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

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

x + y + 2 z = 6

2 x - y + z = 9

y = 2 c o s x

x = \frac{π}{4}

Join BYJU'S Learning Program