Question

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

• A. 147
• B. 143
• C. 137
• D. 157

Answer 1

Answer: (D)

157

Answer 2

Solution Explanation

1. Rewrite the Ellipse
   The ellipse is given by

$$9x^2+4y^2=36 \quad\Rightarrow\quad \frac{x^2}{4}+\frac{y^2}{9}=1.$$

2. Chord through P
   Given $P\left(\frac{2\sqrt{3}}{\sqrt{7}},\,\frac{6}{\sqrt{7}}\right)$. Its line through the origin has slope

$$m=\frac{6/\sqrt{7}}{2\sqrt{3}/\sqrt{7}}=\sqrt{3}.$$

Substitute $y=\sqrt{3}x$ in the ellipse:

$$\frac{x^2}{4}+\frac{3x^2}{9}=\frac{x^2}{4}+\frac{x^2}{3}=\frac{7x^2}{12}=1\quad\Rightarrow\quad x^2=\frac{12}{7}.$$

Thus, length $PQ= 2\sqrt{\frac{12}{7}}=\frac{8\sqrt{3}}{\sqrt{7}}$ and

$$(PQ)^2=\frac{192}{7}\quad\Rightarrow\quad \frac{1}{(PQ)^2}=\frac{7}{192}.$$

3. Perpendicular Chord RS
   Since $RS$ is perpendicular to $PQ$, its slope $m'=-\frac{1}{\sqrt{3}}$. Using $y=-\frac{1}{\sqrt{3}}x$ in the ellipse:

$$\frac{x^2}{4}+\frac{x^2/3}{9}=\frac{x^2}{4}+\frac{x^2}{27}=\frac{31x^2}{108}=1\quad\Rightarrow\quad x^2=\frac{108}{31}.$$

So, chord $RS$ has endpoints $\left(\pm\sqrt{\frac{108}{31}},\, \mp\frac{\sqrt{\frac{108}{31}}}{\sqrt{3}}\right)$. Its length is

$$RS=2\sqrt{\frac{108}{31}}=\frac{24}{\sqrt{31}},$$

giving

$$(RS)^2=\frac{576}{31}\quad\Rightarrow\quad \frac{1}{(RS)^2}=\frac{31}{576}.$$

4. Sum the Reciprocals

$$\frac{1}{(PQ)^2}+\frac{1}{(RS)^2}=\frac{7}{192}+\frac{31}{576}=\frac{21+31}{576}=\frac{52}{576}=\frac{13}{144}.$$

Here, $p=13$ and $q=144$ (coprime), so

$$p+q=13+144=157.$$

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Answer:

$$\boxed{157}\quad \text{(Option 34\% 157)}$$

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Subject, Chapter, and Topic:

• Subject: Mathematics
• Chapter: Conic Sections
• Topic: Ellipses

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Difficulty Level: Medium
Question Type: single_choice