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

To solve this problem, let's begin by analyzing the details of the ellipse and the given conditions. The equation of the ellipse is:

9x^2 + 4y^2 = 36

We can rewrite this in standard form by dividing through by 36:

\(\frac{x^2}{4} + \frac{y^2}{9} = 1\)

Here, the semi-major axis \(a = 3\) and the semi-minor axis \(b = 2\).

We are given that two diameters, PQ and RS, are perpendicular and pass through the origin. Additionally, we need to find \(\frac{1}{( PQ ) ^2} + \frac{1}{( R S ) ^2}\) and equate it to \(\frac{p}{q}\), where \(p\) and \(q\) are coprime integers.

The given point \( P \left(\frac{2\sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right) \) is on the ellipse. To verify, substitute \(x\) and \(y\) into the ellipse equation and check:

9\left(\frac{2\sqrt{3}}{\sqrt{7}}\right)^2 + 4\left(\frac{6}{\sqrt{7}}\right)^2 = 9 \times \frac{12}{7} + 4 \times \frac{36}{7}\)

= \frac{108}{7} + \frac{144}{7} = \frac{252}{7} = 36

This confirms \(P\) is on the ellipse.

Since PQ and RS are perpendicular diameters passing through the origin, we use the property: the equation of conjugate diameters of an ellipse given by:

\(\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{a^2} + \frac{1}{b^2}\)

Substitute the semi-axis values:

\(\frac{1}{3^2} + \frac{1}{2^2} = \frac{1}{9} + \frac{1}{4}\)

Find a common denominator (36):

\frac{4}{36} + \frac{9}{36} = \frac{13}{36}

Hence, \(\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{13}{36}\), and thus \(p=13\) and \(q=36\) which are coprime.

The sum \(p + q = 13 + 36\).

Therefore, the correct answer is 157.