Question

Let \(P(\frac{2√3}{√7} ,\frac{ 6}{√7} ), \)Q,R and S be four points on the ellipse 9x² + 4y² = 36. Let PQ and RS be mutually perpendicular and pass through the origin. If  \(\frac{1}{( P Q ) ^2} + \frac{1}{( R S ) ^2} = \frac{P}{q }\),   where p and q are coprime, then p + q is equal to

• A. 137
• B. 143
• C. 147
• D. 157

Answer 1

The Correct Option is D

Let \( R(2\cos\theta, 3\sin\theta) \) represent the point such that \( OP \perp OR \).

To verify the condition of perpendicularity, we calculate the dot product of the direction ratios of \( OP \) and \( OR \), which should equal -1:

\( 3\sin\theta \times \frac{6}{\sqrt{7}} + 2\cos\theta \times \frac{2\sqrt{3}}{\sqrt{7}} = -1 \)

Simplify the equation:

\( \Rightarrow \tan\theta = -\frac{2}{3\sqrt{3}} \)

Using the value of \( \tan\theta \), the possible coordinates of \( R \) are determined as:

\( R\left( -\frac{6\sqrt{3}}{\sqrt{31}}, \frac{6}{\sqrt{31}} \right) \text{ or } R\left( \frac{6\sqrt{3}}{\sqrt{31}}, -\frac{6}{\sqrt{31}} \right) \)

Now, we compute the required expression for the sum of reciprocals of the squares of the distances:

\( \frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4} \left[ \frac{1}{(OP)^2} + \frac{1}{(OR)^2} \right] \)

Substitute the values for \( (OP)^2 \) and \( (OR)^2 \) based on the given coordinates:

\( = \frac{1}{4} \left[ \frac{1}{\frac{48}{7}} + \frac{1}{\frac{144}{31}} \right] \)

Combine the terms under a common denominator:

\( = \frac{1}{4} \left[ \frac{7}{48} + \frac{31}{144} \right] \)

Perform the calculations:

\( = \frac{1}{4} \left[ \frac{21+31}{144} \right] = \frac{1}{4} \left[ \frac{52}{144} \right] = \frac{13}{144} \)

Thus, the value of \( p + q \) is given as:

Quick Tip

\( p + q = 13 + 144 = 157 \)