Question

The curve which satisfies the differential equation $y^{\prime }=\frac{3x}{y}$ (where y' denotes the first order derivative of y with respect to x) and passes through (1,1) is:

• A. a pair of lines passing through (0,0)
• B. a hyperbola with eccentricity 2
• C. a hyperbola with eccentricity $\frac{2}{\sqrt{3}}$
• D. an ellipse with eccentricity $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B a hyperbola with eccentricity 2

$y^{\prime }=\frac{3x}{y}\Rightarrow \frac{dy}{dx}=\frac{3x}{y}\Rightarrow ydy=3xdx$

Integrating both sides

$\int ydy=\int 3xdx\Rightarrow \frac{y^2}{2}=\frac{3x^2}{2}+c\Rightarrow y^2-3x^2=C$

As it passes through $(1,1)$

$\therefore 1-3=C\Rightarrow C=-2$

Hence $\frac{3x^2}{2}-\frac{y^2}{2}=1$

This is a hyperbola where

$a^2=\frac{2}{3},b^2=2\Rightarrow e^2=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{2.3}{2}}=2$