The curve x^4-2xy^2+y^2+3x-3y=0 cuts the X-ax | vedclass

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(A) The given equation of the curve is $x^4-2xy^2+y^2+3x-3y=0 \dots (i)$.To find the angle at which the curve cuts the $X$-axis,we need to find the sl

Vedclass · Accepted Answer

$\frac{\pi}{4}$

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(A) The given equation of the curve is $x^4-2xy^2+y^2+3x-3y=0 \dots (i)$.To find the angle at which the curve cuts the $X$-axis,we need to find the slope of the tangent at the point $(0,0)$.Differentiating equation $(i)$ with respect to $x$,we get:$4x^3 - 2(y^2 + x \cdot 2y \frac{dy}{dx}) + 2y \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.Simplifying the expression:$4x^3 - 2y^2 - 4xy \frac{dy}{dx} + 2y \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.Now,substitute $(x,y) = (0,0)$ into the derivative equation:$4(0)^3 - 2(0)^2 - 4(0)(0) \frac{dy}{dx} + 2(0) \frac{dy}{dx} + 3 - 3 \frac{dy}{dx} = 0$.$0 - 0 - 0 + 0 + 3 - 3 \frac{dy}{dx} = 0$.$3 = 3 \frac{dy}{dx} \implies \frac{dy}{dx} = 1$.Since the slope $m = 	an 	heta = 1$,we have $	heta = 	an^{-1}(1) = \frac{\pi}{4}$.

The curve $x^4-2xy^2+y^2+3x-3y=0$ cuts the $X$-axis at $(0,0)$ at an angle of

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