Question

If a curve passing through $(1,1)$ is such that the tangent drawn at any point P on it intersects the x-axis at Q and the reciprocal of abscissa of point P is equal to twice x-intercept of a 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. $2y^2-x^2=1$

Answer 1

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

let circle $y=f\left(x\right)$

tangent at $P(x,y)$ at the $x$ axis

at $Q(x-y\frac{dy}{dx},0)$

given is

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

$2y\frac{dy}{dx}=x$

$\int 2ydy=\int xdx$

$y^2=\frac{x^2}{2}+C$

curve passe through $(1,1)$

$c=\frac{1}{2}$

so curve is

$y^2=\frac{x^2}{2}+\frac{1}{2}$

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