This problem is a fantastic illustration of how polar coordinates can simplify geometric problems involving ellipses, especially when distances from the origin and angles are central to the question. The core idea is to express points on the ellipse in terms of their distance from the origin ($r$) and the angle they make with the x-axis ($\phi$).

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1. Standardizing the Ellipse Equation and Deriving its Polar Form

The first crucial step in any ellipse problem is to convert the given equation into its standard form. This helps us identify the semi-major and semi-minor axes, which are fundamental to understanding the ellipse's shape and dimensions.

The given equation is: $9 x^{2}+4 y^{2}=36$ To get it into the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (or vice-versa), we divide the entire equation by the constant term on the right-hand side, which is 36: $\frac{9x^2}{36} + \frac{4y^2}{36} = \frac{36}{36}$ $\frac{x^2}{4} + \frac{y^2}{9} = 1$ Now, comparing this with the standard form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (since $9 > 4$, the major axis is along the y-axis):

• $b^2 = 4 \implies b = 2$. This is the semi-minor axis length.
• $a^2 = 9 \implies a = 3$. This is the semi-major axis length.

Next, we derive the polar form of this ellipse equation. This form is particularly useful because it directly relates the distance of a point from the origin ($r$) to the angle ($\phi$) that the line segment connecting the point to the origin makes with the positive x-axis. We substitute $x = r \cos\phi$ and $y = r \sin\phi$ into the standard ellipse equation: $\frac{(r \cos\phi)^2}{4} + \frac{(r \sin\phi)^2}{9} = 1$ $r^2 \left( \frac{\cos^2\phi}{4} + \frac{\sin^2\phi}{9} \right) = 1$ Rearranging this equation to solve for $\frac{1}{r^2}$ provides a powerful formula that will be central to solving this problem: $\frac{1}{r^2} = \frac{\cos^2\phi}{4} + \frac{\sin^2\phi}{9} \quad \text{--- (General Polar Form Equation)}$ This equation tells us the reciprocal of the square of the distance from the origin to any point on the ellipse at an angle $\phi$.

2. Analyzing the Line Segment PQ

The problem states that P, Q, R, S are points on the ellipse. The lines PQ and RS pass through the origin and are mutually perpendicular.

Let's consider point P, given as $\left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right)$. Let $r_P$ be the distance of point P from the origin $O(0,0)$. We calculate $r_P^2$ using the distance formula: $r_P^2 = \left(\frac{2 \sqrt{3}}{\sqrt{7}}\right)^2 + \left(\frac{6}{\sqrt{7}}\right)^2$ $r_P^2 = \frac{(2\sqrt{3})^2}{7} + \frac{6^2}{7} = \frac{12}{7} + \frac{36}{7} = \frac{48}{7}$ Therefore, $\frac{1}{r_P^2} = \frac{7}{48}$.

Since the line segment PQ passes through the origin, P and Q must be diametrically opposite points on the ellipse. This means that the distance from the origin to Q ($OQ$) is equal to the distance from the origin to P ($OP$). So, $OQ = OP = r_P$. The total length of the chord PQ is $OP + OQ = r_P + r_P = 2r_P$. Therefore, $(PQ)^2 = (2r_P)^2 = 4r_P^2$. From this, we get $\frac{1}{(PQ)^2} = \frac{1}{4r_P^2}$.

Now, let $\theta$ be the angle that the line OP (and thus the line PQ) makes with the positive x-axis. We can find $\cos\theta$ and $\sin\theta$ using the coordinates of P and $r_P$. The coordinates of P are $(x_P, y_P) = (r_P \cos\theta, r_P \sin\theta)$. We have $r_P = \sqrt{\frac{48}{7}} = \frac{\sqrt{16 \times 3}}{\sqrt{7}} = \frac{4\sqrt{3}}{\sqrt{7}}$. So, $\cos\theta = \frac{x_P}{r_P} = \frac{2\sqrt{3}/\sqrt{7}}{4\sqrt{3}/\sqrt{7}} = \frac{1}{2}$ $\sin\theta = \frac{y_P}{r_P} = \frac{6/\sqrt{7}}{4\sqrt{3}/\sqrt{7}} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$ Thus, $\cos^2\theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$ and $\sin^2\theta = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$.

Self-check: We can verify these values using our General Polar Form Equation: $\frac{1}{r_P^2} = \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9} = \frac{1/4}{4} + \frac{3/4}{9} = \frac{1}{16} + \frac{3}{36} = \frac{1}{16} + \frac{1}{12}$ To sum these fractions, find a common denominator (48): $\frac{3}{48} + \frac{4}{48} = \frac{7}{48}$ This matches our direct calculation of $\frac{1}{r_P^2} = \frac{7}{48}$, confirming our values for $\cos^2\theta$ and $\sin^2\theta$ are correct.

Now, we can express $\frac{1}{(PQ)^2}$ in terms of $\theta$: $\frac{1}{(PQ)^2} = \frac{1}{4r_P^2} = \frac{1}{4} \left( \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9} \right) \quad \text{--- (Equation A)}$

3. Analyzing the Line Segment RS

The problem states that the line RS also passes through the origin and is mutually perpendicular to PQ. If the line PQ makes an angle $\theta$ with the positive x-axis, then a line perpendicular to it must make an angle of $\theta + \frac{\pi}{2}$ (or $\theta - \frac{\pi}{2}$) with the positive x-axis. Let's use $\phi' = \theta + \frac{\pi}{2}$ for the angle of line RS.

Let $r_R$ be the distance of point R from the origin. Using our General Polar Form Equation for point R (at angle $\phi'$): $\frac{1}{r_R^2} = \frac{\cos^2\phi'}{4} + \frac{\sin^2\phi'}{9}$ Substitute $\phi' = \theta + \frac{\pi}{2}$: $\frac{1}{r_R^2} = \frac{\cos^2\left(\theta + \frac{\pi}{2}\right)}{4} + \frac{\sin^2\left(\theta + \frac{\pi}{2}\right)}{9}$ Recall the trigonometric identities:

• $\cos\left(\theta + \frac{\pi}{2}\right) = -\sin\theta$
• $\sin\left(\theta + \frac{\pi}{2}\right) = \cos\theta$ Squaring these, we get:
• $\cos^2\left(\theta + \frac{\pi}{2}\right) = (-\sin\theta)^2 = \sin^2\theta$
• $\sin^2\left(\theta + \frac{\pi}{2}\right) = (\cos\theta)^2 = \cos^2\theta$

Substituting these into the equation for $\frac{1}{r_R^2}$: $\frac{1}{r_R^2} = \frac{\sin^2\theta}{4} + \frac{\cos^2\theta}{9}$ Similar to PQ, since RS passes through the origin, R and S are diametrically opposite points on the ellipse. The total length of the chord RS is $OR + OS = r_R + r_R = 2r_R$. Therefore, $(RS)^2 = (2r_R)^2 = 4r_R^2$. From this, we get $\frac{1}{(RS)^2} = \frac{1}{4r_R^2}$. $\frac{1}{(RS)^2} = \frac{1}{4} \left( \frac{\sin^2\theta}{4} + \frac{\cos^2\theta}{9} \right) \quad \text{--- (Equation B)}$

4. Combining the Results to Find the Desired Sum

We need to calculate the value of $\frac{1}{(PQ)^2} + \frac{1}{(RS)^2}$. We simply add Equation A and Equation B: $\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4} \left( \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9} \right) + \frac{1}{4} \left( \frac{\sin^2\theta}{4} + \frac{\cos^2\theta}{9} \right)$ Factor out the common term $\frac{1}{4}$: $= \frac{1}{4} \left[ \left( \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{9} \right) + \left( \frac{\sin^2\theta}{4} + \frac{\cos^2\theta}{9} \right) \right]$ Now, group the terms inside the brackets by $\cos^2\theta$ and $\sin^2\theta$: $= \frac{1}{4} \left[ \cos^2\theta \left( \frac{1}{4} + \frac{1}{9} \right) + \sin^2\theta \left( \frac{1}{9} + \frac{1}{4} \right) \right]$ Notice that the coefficients for $\cos^2\theta$ and $\sin^2\theta$ are the same. Factor out $\left( \frac{1}{4} + \frac{1}{9} \right)$: $= \frac{1}{4} \left[ \left( \frac{1}{4} + \frac{1}{9} \right) (\cos^2\theta + \sin^2\theta) \right]$ We know the fundamental trigonometric identity $\cos^2\theta + \sin^2\theta = 1$. $= \frac{1}{4} \left( \frac{1}{4} + \frac{1}{9} \right) (1)$ Now, perform the addition of the fractions: $\frac{1}{4} + \frac{1}{9} = \frac{9}{36} + \frac{4}{36} = \frac{13}{36}$ Substitute this back: $\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{1}{4} \left( \frac{13}{36} \right) = \frac{13}{144}$

5. Final Calculation and Answer

The problem states that $\frac{1}{(PQ)^2}+\frac{1}{(RS)^2}=\frac{p}{q}$, where $p$ and $q$ are coprime. We found the sum to be $\frac{13}{144}$. Here, $p = 13$ and $q = 144$. Since 13 is a prime number and 144 is $12^2 = (2^4 \times 3^2)$, 13 and 144 share no common factors other than 1. Thus, $p=13$ and $q=144$ are coprime.

Finally, we need to find $p+q$: $p+q = 13 + 144 = 157$

Comparing this with the given options: (A) 143 (B) 147 (C) 137 (D) 157

The calculated value matches option (D).

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Key Takeaways and JEE Tips

1. Polar Form is King for Origin-Centric Problems: When a problem involves distances from the origin (like $OP$, $OQ$) or chords passing through the origin, the polar form of the conic section equation (especially for ellipses and hyperbolas) is often the most efficient approach. Remember the general derivation: $\frac{1}{r^2} = \frac{\cos^2\phi}{b^2} + \frac{\sin^2\phi}{a^2}$ for $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.
2. Diametric Chords: If a chord passes through the origin, the points on the ellipse are diametrically opposite, meaning $PQ = 2 \times OP$. This simple fact