Question

The $x$-intercept of the tangent to a curve is twice the ordinate of the point of contact. The equation of the curve through the point $(1,1)$ is

• A. $y=e^{\frac{y-x}{2y}}$
• B. $x=e^{\frac{2x-y}{2x}}$
• C. $y=e^{\frac{2y-x}{4x}}$
• D. $x=e^{\frac{x-y}{2x}}$

Answer 1

The correct option is A $y=e^{\frac{y-x}{2y}}$

Equation of tangent is $Y-y=\frac{dy}{dx}(X-x)$
for X-intercept $Y=0\Rightarrow X=x-y\frac{dx}{dy}$
Now, $x-y\frac{dx}{dy}=2y$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x-2y}$
Put $y=vx$
$v+x\frac{dv}{dx}=\frac{v}{1-2v}$
$x\frac{dv}{dx}=\frac{2v^2}{1-2v}$
$\int \frac{1-2v}{2v^2}dv=\int \frac{dx}{x}$
$-\frac{1}{2v}-\log v=\log x+c$
$-\frac{x}{2y}-\log \frac{y}{x}=\log x+c$
at $(1,1)\Rightarrow$ $c=-\frac{1}{2}$
$-\frac{x}{2y}-\log \frac{y}{x}=\log x-\frac{1}{2}$
$y=e^{\frac{2y-2x}{4y}}$
$y=e^{\frac{y-x}{2y}}$