Ellipse
Perpendicular Chords Through Origin — Harmonic Sum
nta_pyq_2023_apr
Grade 11
Question:
Let $P(2\sqrt{3},\frac{6}{\sqrt{7}})$, $Q$, $R$ 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 $\dfrac{1}{(PQ)^2}+\dfrac{1}{(RS)^2}=\dfrac{p}{q}$, then $p+q$ is equal to
Step-by-Step Solution
Key Concept: Using the focal chord property for a diameter: if $r_1$ and $r_2$ are semi-lengths of perpendicular diameters of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then $\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}=\frac{1}{r^2}$.
$\frac{1}{PQ^2}+\frac{1}{RS^2}=\frac{13}{144}$. $p+q=13+144=157$.
Correct Answer: 4