Question

Assertion (A): If $PS{P}^1,QS{Q}^1$ are two perpendicular focal chords of the conic $\frac{l}{r}=1+\cos \theta$, then $\frac{1}{SP.S{P}^1}+\frac{1}{SQ.S{Q}^1}$ is constant
Reason (R) : For the conic $\frac{l}{r}=1+e\cos \theta ,\frac{1}{SP}+\frac{1}{SQ}=\frac{2}{l}$ where PSQ is focal chord
The correct answer is

• A. both A and R are true and R is the correct explanation of A
• B. both A and R are true but R is not the correct explanation of A
• C. A is true, R is false
• D. A is false, R is true

Answer 1

The correct option is B both A and R are true but R is not the correct explanation of A

$\frac{l}{SP}=1+\cos \theta \frac{l}{S{P}^1}=1+\cos (\pi +\theta )\frac{l^2}{SP.S{P}^1}=(1+\cos \theta )(1-\cos \theta )={\sin }^2\theta ....\left(1\right)$
Since, $QS{Q}^1$ is perpendicular to $PS{P}^1$, we get
$\frac{l}{QS}=1+\cos (\frac{\pi }{2}+\theta )=1-\sin \theta \frac{l}{S{Q}^1}=1+\cos (\frac{3\pi }{2}+\theta )=1+\sin \theta \frac{l^2}{QS.S{Q}^1}=(1-\sin \theta )(1+\sin \theta )=1-{\sin }^2\theta ={\cos }^2\theta ....\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$ we get
$l^2(\frac{1}{PS.S{P}^1}+\frac{1}{QS.S{Q}^1})={\sin }^2\theta +{\cos }^2\theta =1\Rightarrow \frac{1}{PS.S{P}^1}+\frac{1}{QS.S{Q}^1}=\frac{1}{l^2}$
Hence, Assertion is correct.
Similarly,
If $PSQ$ is a focal conic, we get
$\frac{l}{SP}=(1+\cos \theta )$
$\frac{l}{S{P}^1}=1+\cos (\pi +\theta )=1-\cos \theta$
Adding above two equations, we get
$\frac{l}{SP}+\frac{l}{S{P}^1}=2\Rightarrow \frac{1}{SP}+\frac{1}{S{P}^1}=\frac{2}{l}$
Hence, the statement given in reason is also correct.
But it does not explain the statement in Assertion. Hence, option B is correct.