Question

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

Hint: The two given focal chords of the given conic are said to be perpendicular, so the formula including the equations of the focal chords will be 90$^\circ $ apart, i.e., the difference of the arguments used in the equation must be 90$^\circ $. The particular relations of each of the equations, formulae must be known, for they are the ones which are going to be used in the solving of the question.

Formula Used:
We are going to use some basic trigonometric formula in the question:
${\sin ^2}\theta + {\cos ^2}\theta = 1$, $\cos \left( {\pi + \theta } \right) = - \cos \theta $, $\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \theta $, $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = \sin \theta $
And, if $\dfrac{l}{{SP}} = 1 + \cos \theta $ , then $\dfrac{l}{{S{P^1}}} = 1 + \cos \left( {\pi + \theta } \right)$

Complete step by step solution:
In the given question, we have been given that there are two perpendicular focal chords (which are given as $PS{P^1},{\text{ }}QS{Q^1}$) of the conic which is given to be $\dfrac{l}{r} = 1 + \cos \theta $
So, we can write the above result as follows:
$ \Rightarrow \,\,\dfrac{l}{{SP}} = 1 + \cos \theta $ …..…(i) and
$ \Rightarrow \,\,\,\dfrac{l}{{S{P^1}}} = 1 + \cos \left( {\pi + \theta } \right)$
But we know that $\cos \left( {\pi + \theta } \right) = - \cos \theta $ , so $1 + \cos \left( {\pi + \theta } \right) = 1 - \cos \theta $ and hence, we have got:

$ \Rightarrow \,\,\,\dfrac{l}{{S{P^1}}} = 1 + \cos \left( {\pi + \theta } \right) = 1 - \cos \theta $ ………(ii)
Now, multiplying the two equations (i) and (ii), we get:
$ \Rightarrow \,\,\,\dfrac{l}{{SP}} \times \dfrac{l}{{S{P^1}}} = \left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)$
Here, we can apply the difference of square of the algebra, which is:
$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
Hence, we have got:
$ \Rightarrow \,\,\,\dfrac{l}{{SP}} \times \dfrac{l}{{S{P^1}}} = \left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right) = 1 - {\cos ^2}\theta $
But we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Thus,
$ \Rightarrow \,\,\dfrac{{{l^2}}}{{SP \times S{P^1}}} = \left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right) = {\sin ^2}\theta $ …….…(iii)
Now, since $QS{Q^1}$ is perpendicular to $PS{P^1}$ we get:
$ \Rightarrow \,\,\,\dfrac{l}{{QS}} = 1 + \cos \left( {\dfrac{\pi }{2} + \theta } \right)$
But we know that $\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \theta $, so we get:
$ \Rightarrow \,\,\,\dfrac{l}{{QS}} = 1 + \cos \left( {\dfrac{\pi }{2} + \theta } \right) = 1 - \sin \theta $ …..…(iv) and
$ \Rightarrow \,\,\,\dfrac{l}{{S{Q^1}}} = 1 + \cos \left( {\dfrac{{3\pi }}{2} + \theta } \right)$
We know $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = \sin \theta $, so we have got:
$ \Rightarrow \,\,\,\dfrac{l}{{S{Q^1}}} = 1 + \cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = 1 + \sin \theta $ …….…(v)
Now, multiplying the equations (iii) and (iv), we have:
$ \Rightarrow \,\,\,\dfrac{l}{{QS}} \times \dfrac{l}{{S{Q^1}}} = \left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right) = 1 - {\sin ^2}\theta $
or,
$ \Rightarrow \,\,\,\dfrac{{{l^2}}}{{QS \times S{Q^1}}} = {\cos ^2}\theta $ ………(vi)
Adding the two equations (v) and (vi), we get:
$ \Rightarrow \,\,\,\dfrac{{{l^2}}}{{SP.S{P^1}}} + \dfrac{{{l^2}}}{{SQ.S{Q^1}}} = {\sin ^2}\theta + {\cos ^2}\theta $
or,
$ \Rightarrow \,\,\,\dfrac{1}{{SP.S{P^1}}} + \dfrac{1}{{SQ.S{Q^1}}} = \dfrac{1}{{{l^2}}}$, which is a constant
Hence, Assertion (A) is correct.
Now, if $PSQ$ is a focal chord of the given conic, we have:
$ \Rightarrow \,\,\dfrac{l}{{SP}} = \left( {1 + \cos \theta } \right)$
$ \Rightarrow \,\,\dfrac{l}{{S{P^1}}} = 1 + \left( {\cos \pi + \theta } \right) = 1 - \cos \theta $
Adding the two equations, we get:
$ \Rightarrow \,\,\dfrac{l}{{SP}} + \dfrac{l}{{S{P^1}}} = 2$
or,
$ \Rightarrow \,\,\dfrac{1}{{SP}} + \dfrac{1}{{S{P^1}}} = \dfrac{2}{l}$
Hence, Reason (R) is correct, but it is not a correct explanation of Assertion (A).
So, the correct answer is Option B.

Note: So, we saw that in solving questions like these, it is very important that we know the exact relations, formulae, concepts of the topic of the question, as any half-known thing, or something with incomplete knowledge yields incorrect results.