Question

A curve passing through point $(1,2)$ possessing the following property; the segment of the tangent between the point of tangency & the x-axis is bisected at the point of intersection with the y-axis. If $A$ is area bounded by the curve & line $x=1$ then $9A^2$ is equal to

Answer 1

Let $y=f\left(x\right)$ be the curve,

Now, equation of the tangent at $P(x,y)$ to the curve is

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

or, $Y-X{y}^{\prime }=y-x{y}^{\prime }$

or, $\frac{X}{\frac{x{y}^{\prime }-y}{y^{\prime }}}+\frac{Y}{y-x{y}^{\prime }}$ $=1$.

The tangent cuts the $x$-axis and $y-$ axis at $A\left(\frac{x{y}^{\prime }-y}{y^{\prime }},0\right)$ and $B(0,y-x{y}^{\prime })$.

According to the problem

$\frac{x+\frac{x{y}^{\prime }-y}{y^{\prime }}}{2}$ $=0$ and $\frac{y}{2}$ $=y-x{y}^{\prime }$.

From above two equations we get the same equation

$2x{y}^{\prime }=y$

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

Integrating we get,

$\log y^2=\log cx$ [$c$ being the integrating constant].

or, $y^2=cx$.

This curve passes through $(1,2)$ $\Rightarrow c=4$.

So, the curve is $y^2=4x$.

Now the area bounded by the curve and $x=1$ is shown in the figure and it is $=2\underset{0}{\overset{1}{\int }}2\sqrt{x}dx=\frac{8}{3}$.

By the problem $A=\frac{8}{3}$.

$\therefore 9A^2=64$.