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If the slope of the tangent to the curve y=y(x)y = y(x) at any point (x,y)(x, y) is 2xy2\frac{2x}{y^2} and the curve passes through the point (13,1)\left(\frac{1}{\sqrt3}, 1\right), then equation of curve is

Solution

Correct Option: 2

The slope of the tangent at any point (x,y)(x, y) on the curve is dydx\dfrac{dy}{dx}, so:

dydx=2xy2\dfrac{dy}{dx} = \dfrac{2x}{y^2}


y2dy=2xdxy^2 \, dy = 2x \, dx

y2dy=2xdx\displaystyle\int y^2 \, dy = \int 2x \, dx

y33=2x22+C\dfrac{y^3}{3} = \dfrac{2x^2}{2} + C

y33=x2+C\dfrac{y^3}{3} = x^2 + C


The curve passes through (13, 1)\left(\dfrac{1}{\sqrt{3}},\ 1\right), so substituting x=13x = \dfrac{1}{\sqrt{3}} and y=1y = 1:

(1)33=(13)2+C\dfrac{(1)^3}{3} = \left(\dfrac{1}{\sqrt{3}}\right)^2 + C

13=13+C\dfrac{1}{3} = \dfrac{1}{3} + C

C=0C = 0


y33=x2\dfrac{y^3}{3} = x^2

y3=3x2\boxed{y^3 = 3x^2}

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