The two curves x3-3 x y2-2-0 and 3 x2 y-y3-2A- touch each otherB- cut at right angleC- cut at angle -3D- cut at angle 4

X^3 3xy^2+2=0 nbsp; ⇒dydx=x^2 y^22xy=m_1 (say) 3x^2y y^3=2 ⇒dydx= 2xyx^2 y^2=m_2 (say) Product of the slopes at any point (x,y)=m_1 × m_2= 1 Therefore, the c ...

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Standard XII

Mathematics

Critical Point

The two curve...

Question

The two curves $x^{3} - 3 x y^{2} + 2 = 0$ and $3 x^{2} y - y^{3} = 2$

touch each other

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cut at right angle

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cut at angle

\frac{π}{3}

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cut at angle

\frac{π}{4}

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Solution

The correct option is B cut at right angle
$x^{3} - 3 x y^{2} + 2 = 0$
$\Rightarrow \frac{d y}{d x} = \frac{x^{2} - y^{2}}{2 x y} = m_{1} (say)$

$3 x^{2} y - y^{3} = 2$
$\Rightarrow \frac{d y}{d x} = \frac{- 2 x y}{x^{2} - y^{2}} = m_{2} (say)$
Product of the slopes at any point $(x, y) = m_{1} \times m_{2} = - 1$
Therefore, the curves are orthogonal and cut each other at right angle.

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Similar questions

Q. Two curves x³ - 3xy² + 2 = 0 and 3x²y - y³ = 2
(a) touch each other (b) cut at right angle
(c) cut an angle

\frac{π}{3}

(d) cut at an angle

\frac{π}{4}

Q. The two curves

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

and

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

The two curves $x^{3} - 3 x y^{2} + 2 = 0$ and $3 x^{2} y - y^{3} = 2$

Q. The two curves

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

and

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

intersect at an angle of?

Q. An angle at which the two curves

x^{3} - 3 x y^{2} = 4

and

3 x^{2} y - y^{3} = 4

intersect is

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The two curves x3−3xy2+2=0 and 3x2y−y3=2

The correct option is B cut at right angle x3−3xy2+2=0 ⇒dydx=x2−y22xy=m1 (say) 3x2y−y3=2 ⇒dydx=−2xyx2−y2=m2 (say) Product of the slopes at any point (x,y)=m1×m2=−1 Therefore, the curves are orthogonal and cut each other at right angle.

The two curves $x^{3} - 3 x y^{2} + 2 = 0$ and $3 x^{2} y - y^{3} = 2$