Question

If a curve passing through $(1,1)$ is such that the tangent drawn at any point $P$ in it intersects the x-axis at $Q$ and the recriprocal of abscissa of point $P$ is equal to twice x-intercept of tangent at $P$. Then the equation of the curve is

• A. $y^2=x$
• B. $x^2=2y-1$
• C. $3x^2-2y^2=1$
• D. $2x^2-y^2=1$

Answer 1

The correct option is D $2x^2-y^2=1$

$2(x-\frac{y}{\frac{dy}{dx}})=\frac{1}{x}$

$x-\frac{1}{2x}=\frac{y}{\frac{dy}{dx}}\Rightarrow \frac{(2x^2-1)}{2x}=\frac{y}{\frac{dy}{dx}}$

$\int \frac{dy}{y}=\int \frac{2x}{(2x^2-1)}dx$

so, $y=k{(2x^2-1)}^{1/2}$

Here $k=1$, thus $y^2=2x^2-1$

$2x^2-y^2=1$