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 which of the following is/are correct?

• A. Focus of $C$ is $(4,0)$.
• B. Length of latus rectum of $C$ is $4$ units.
• C. The area bounded by curve $C$ and its latus rectum is equal to $\frac{8}{3}$ sq. units.
• D. The locus of mid points of focal chords of curve $C$ is $y^2=2(x-1)$

Answer 1

Answer: (B), (C), (D)

The locus of mid points of focal chords of curve $C$ is $y^2=2(x-1)$

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 $=(1,0)$
Length of latus rectum $=4$ units


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

Let the endpoints of the focal chord be $A=(t_1^2,2t_1)$ and $B=(t_2^2,2t_2)$
So, $t_1\times t_2=-1$
Let the midpoint of $A$ and $B$ be $M(h,k)$, then
$2h=t_1^2+t_2^{2}k=(t_1+t_2)$
Now, $(t_1+t_2{)}^2=t_1^2+t_2^2+2t_1{t}_2$
$\Rightarrow k^2=2h-2$
Hence, the locus is $y^2=2(x-1)$