Question

The slope of the tangent at (x,y) to a curve passing through a point (2,1) is $\frac{x^{2} + y^{2}}{2xy}$ then the equation of the curve is

• A. $2(x^{2} - y^{2}) = 3x$
• B. $2(x^{2} - y^{2}) = 6x$
• C. $x(x^{2} - y^{2}) = 6$
• D. $x(x^{2} + y^{2}) = 10$

Answer 1

Answer: (A)

$2(x^{2} - y^{2}) = 3x$

Answer 2

$\frac{dy}{dx} = \frac{x^{2} + y^{2}}{2xy}.$ Put $y = \nu x \Rightarrow \nu + x.\frac{d\nu}{dx} = \frac{s^{2} + \nu^{2}x^{2}}{2\nu x^{2}}$

$\frac{2\nu}{1 - \nu^{2}}.d\nu = \frac{dx}{x}$

Integrating both sides, - log(1 - v2) = log x + log c

$- \log\left( 1 - \frac{y^{2}}{x^{2}} \right) = \log x + \log c$

This passes through (2,1)

$- \log\left( 1 - \frac{1}{4} \right) = \log 2 + \log c \Rightarrow c = \frac{2}{3}$

From equation (i) $\log\left( \frac{x^{2}}{x^{2} - y^{2}} \right)$ = log xc

2(x2 - y2) = 3x.