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JEE Advanced 2014
Paper-2 2014
Question
Let a, r, s, t be nonzero real numbers. Let $P\left(a t^2, 2 a t\right), Q, R\left(a r^2, 2 a r\right)$ and $\left(a s^2, 2 a s\right)$ be distinct points on the parabola $\mathrm{y}^2=4 \mathrm{ax}$. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a, 0)
The value of r is
Select the correct option:
A
$-\frac{1}{t}$
B
$\frac{t^2+1}{t}$
C
$\frac{1}{t}$
D
$\frac{t^2-1}{t}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & m_{P K}=m_{Q R} \\ & \frac{2 a t-0}{a t^2-2 a}=\frac{2 a t^{\prime}-2 a r}{a\left(t^{\prime}\right)^2-a r^2} \\ & \frac{t}{t^2-2}=\frac{t^{\prime}-r}{\left(t^{\prime}\right)^2-r^2} \\ & -t^{\prime}-t r^2=-t-r t^2-2 t^{\prime}+2 r, \quad t t^{\prime}=-1 \\ & t^{\prime}-t r^2=-t+2 r-r t^2 \\ & -t r^2+r\left(t^2-2\right)+t^{\prime}+t=0 \\ & \lambda=\frac{\left(2-t^2\right) \pm \sqrt{\left(t^2-2\right)^2+4\left(-1+t^2\right)}}{-2 t} \\ & =\frac{\left(2-t^2\right) \pm \sqrt{t^4}}{-2 t}=\frac{2-t^2 \pm t^2}{-2 t}\end{aligned}$
$r=-\frac{1}{t}$ It is not possible as the $R$ \& $Q$ will be one same.
or $r=\frac{t^2-1}{t}$
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