Q94 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics
If $P Q$ is a focal chord of the parabola $y^2=4 x$ with focus $S$ and $P=(4,4)$, then $S Q=$
- A2
- Correct answerB$\frac{5}{4}$
- C5
- D$\frac{3}{2}$
Solution
$P Q$ is focal chord of parabola $y^2=4 x$ focus $S(1,0)$ and $P(4,4)$
We know that, Length of Semi-latus rectum is Harmonic mean of $P S$ and $S Q$

$\begin{aligned} \therefore \quad \frac{1}{P S}+\frac{1}{S Q} & =1 \Rightarrow \frac{1}{S Q}=1-\frac{1}{P S} \\ \frac{1}{S Q} & =1-\frac{1}{\sqrt{9+16}}=1-\frac{1}{5}=\frac{4}{5} \Rightarrow S Q=\frac{5}{4}\end{aligned}$

$\begin{aligned} \therefore \quad \frac{1}{P S}+\frac{1}{S Q} & =1 \Rightarrow \frac{1}{S Q}=1-\frac{1}{P S} \\ \frac{1}{S Q} & =1-\frac{1}{\sqrt{9+16}}=1-\frac{1}{5}=\frac{4}{5} \Rightarrow S Q=\frac{5}{4}\end{aligned}$
Hence, the correct answer is option: 'B'.