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MHTCET 10 May 2024 Shift 1 Paper

Question : 108 of 150

Marks: +1, -0

The curve

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

(0,0)

at an angle of

Solution:

Concept:

The angle at which a curve cuts the X-axis is the angle made by its tangent with the positive X-axis.

Explanation:

Given curve is:

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

Differentiating both sides with respect to

x

4x^{3}-2y^{2}+3+\left(-4xy+2y-3\right)\frac{dy}{dx}=0

So, the slope of the tangent is:

m=\frac{dy}{dx}=\frac{4x^{3}-2y^{2}+3}{4xy-2y+3}

At the point

(0,0)

m=\frac{4(0)-2(0)+3}{4(0)(0)-2(0)+3}=\frac{3}{3}=1

Since slope

m=\tan \theta

, we get:

\tan \theta=1

Therefore:

\theta=\frac{\pi}{4}

Answer:

The curve cuts the X-axis at

(0,0)

at an angle of

\frac{\pi}{4}

Hence, the correct option is A.

\frac{\pi}{4}

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