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The angle between intersection of curves $x^3 - 3xy^2 = 4$ and $3x^2y - y^3 = 4$ is

Option 1)
$\frac{\pi}{4}$

Option 2)
$\frac{\pi}{3}$

Option 3)
$\frac{\pi}{}2$

Option 4)
$\frac{3\pi}{4}$

Answers (1)

best_answer

As we have learned

Condition of Orthogonality -

Two curves intersect each other orthogonally if the tangents to each of them subtend a right angle at the point of intersection of two curves:

$m_{1}\times m_{2}=-1$

-

Let (x,y) is point of intersection

$x^{3}-3xy^{2}=4 \Rightarrow 3x^{2} -3[x(2y)\frac{dy}{dx}+y^{2}]=0$

$\Rightarrow \frac{dy}{dx} = \frac{x^{2}-y^{2}}{2xy }$

$\therefore \frac{dy}{dx}|_{(x,y )}=$ slope of tangent at (x,y) = $= \frac{x_{1}^{2}-y_{1}^{2}}{2x_{1}y_{1} }$

$3x^{2}y-y^{3}\Rightarrow 3x^{2}\frac{dy}{dx} + 6xy -3y^{2}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx} =-2xy/ x^{2}-y^{2}$

$\therefore \frac{dy}{dx} |_{(x,y)} = \frac{-2x_{1}y_{1}}{x_{1}^{2}-y_{1}^{2}}=$ slope of tangent at (x,y )

Now $\therefore \frac{dy}{dx} |_{(x,y)}$ of first $\times \frac{d}{dx}|_{(x,y)}$ of second

$= \frac{x_{1}^{2}-y_{1}^{2}}{2x_{1}y_{1} }\times$ $\frac{-2x_{1}y_{1}}{x_{1}^{2}-y_{1}^{2}}=-1$

it is orthogonal intersection , so angle $=\pi /2$

Option 1)

$\frac{\pi}{4}$

Option 2)

$\frac{\pi}{3}$

Option 3)

$\frac{\pi}{}2$

Option 4)

$\frac{3\pi}{4}$

Posted by

Himanshu

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