Question

Let $P\left(t_1\right),Q\left(t_2\right)$ be two points on the parabola $y^2=8x.$ Then which of the following is/are true?

• A. If the chord $PQ$ meets the $x-$axis at $(4,0)$, then $t_1{t}_2=-2$
• B. If $PQ$ is a focal chord, then $t_1{t}_2=-2$
• C. If $t_1=2$, then the length of focal chord $PQ$ is $\frac{25}{4}$
• D. If $t_1=-\frac{1}{2}$ and $PQ$ meets the $x-$axis at $(4,0)$, then $Q\equiv (32,16)$

Answer 1

Answer: (A), (D)

If $t_1=-\frac{1}{2}$ and $PQ$ meets the $x-$axis at $(4,0)$, then $Q\equiv (32,16)$

$P\left(t_1\right)$ and $Q\left(t_2\right)$ are two points on the parabola $y^2=8x$
Here, $a=2$

Equation of chord $PQ$ is $(t_1+t_2)y=2x+2a{t}_1{t}_2$
If this chord meets the $x-$axis at $(c,0)$, then
$0=2c+2a{t}_1{t}_2$
$\Rightarrow t_1{t}_2=-\frac{c}{2}(\because a=2)$

If $PQ$ meets the $x-$axis at $(4,0)$, then
$t_1{t}_2=-\frac{4}{2}=-2$

If $PQ$ is a focal chord that is passing through $(2,0)$, then
$t_1{t}_2=-\frac{2}{2}=-1$

If $t_1=2$ and $PQ$ is focal chord, then length of focal chord $=a{(t_1+\frac{1}{t_1})}^2=2\times \frac{25}{4}=\frac{25}{2}$

If $t_1=-\frac{1}{2}$ and $PQ$ meets the $x-$axis at $(4,0)$, then
$t_1{t}_2=-\frac{4}{2}=-2$
$\Rightarrow t_2=-\frac{2}{t_1}=4$
$\therefore Q\equiv (a{t}_2^2,2a{t}_2)\equiv (32,16)$