Question

Let $P(a{t}^2,2at),Q(a{r}^2,2ar)$ be three points on a parabola $y^2=4ax$. If $PQ$ is the focal chord and $PK,QR$ are parallel where the co-ordinates of $K$ is $(2a,0)$, then the value of $r$ is

• A. $\frac{t}{1-t^2}$
• B. $\frac{1-t^2}{t}$
• C. $\frac{t^2+1}{t}$
• D. $\frac{t^2-1}{t}$

Answer 1

The correct option is D $\frac{t^2-1}{t}$

$m_{PK}=m_{QR}$
$\frac{2at-0}{a{t}^2-2a}=\frac{2at-2ar}{a{\left(t^{\prime }\right)}^2-a{r}^2}$
$-t^{\prime }-t{r}^2=-t-r{t}^2-2t^{\prime }+2r,t{t}^{\prime }=-1$
$t^{\prime }-t{r}^2=-t+2r-r{t}^2$
$-t{r}^2+r(t^2-2)+t^{\prime }+t=0$
$\lambda =\frac{(2-t^2)\pm \sqrt{{(t^2-2)}^2+4(-1+t^2)}}{-2t}=\frac{2-t^2\pm t^2}{-2t}$
$r=-\frac{1}{t}$
It is not possible as the $R$ and $Q$ will be one same
or $r=\frac{t^2-1}{t}$