Question

A family of curves has the property that the segment of the tangent between the point of tangency and $x-$axis is bisected at the point of intersection with $y-$axis. If a member of this family $C$ passes through the point $(9,–6)$, then the area bounded by curve $C$ and its latus rectum is equal to

• A. $\frac{8}{3}$ sq. units
• B. $\frac{4}{3}$ sq. units
• C. $\frac{2}{3}$ sq. units
• D. $\frac{16}{3}$ sq. units

Answer 1

The correct option is A $\frac{8}{3}$ sq. units

Equation of tangent at $P(x,y)$ is
$(Y-y)=\frac{dy}{dx}(X-x)$
When $Y=0$
$X=x-\frac{y}{dy/dx}\Rightarrow T=(x-\frac{y}{dy/dx},0)$
When $X=0$
$Y=y-\frac{dy}{dx}x\Rightarrow M=(0,y-\frac{dy}{dx}x)$


As $M$ is the mid point of $P$ and $T$, so
$(0,y-\frac{dy}{dx}x)=(\frac{x+x-\frac{y}{dy/dx}}{2},\frac{y}{2})\Rightarrow x-\frac{y/2}{dy/dx}=0,y-\frac{dy}{dx}x=\frac{y}{2}\Rightarrow \frac{dy}{dx}=\frac{y}{2x},\frac{dy}{dx}=\frac{y}{2x}\Rightarrow 2\int \frac{dy}{y}=\int \frac{dx}{x}+C\Rightarrow 2\ln |y|=\ln |x|+C$
The curve passes through $(9,-6)$, then
$2\ln 6=\ln 9+C\Rightarrow C=\ln 4\Rightarrow |y{|}^2=4|x|$
As this passes through $(9,-6)$, so
$y^2=4x\Rightarrow a=1$
Focus of the parabola $=(1,0)$
Endpoints of latus rectum are $(1,2)$ and $(1,-2)$


Area bounded by $C$ and its latus rectum
$=2\underset{0}{\overset{1}{\int }}\sqrt{4x}dx$
$=4{\left[\frac{2}{3}{x}^{3/2}\right]}_0^1=\frac{8}{3}$ sq. units