Coordinate Geometry
Ellipse
MMTS_Full_Test_09
Grade 12
Question:
Let $P\left(\frac{2\sqrt{3}}{\sqrt{7}},\frac{6}{\sqrt{7}}\right)$, $Q$, $R$ and $S$ be four points on ellipse $9x^2+4y^2=36$. Let $PQ$ and $RS$ be mutually perpendicular chords and pass through the centre of ellipse. Then $\left[\frac{50}{PQ^2}+\frac{50}{RS^2}\right]=$
Step-by-Step Solution
Key Concept: For conjugate diameters of ellipse $x^2/a^2+y^2/b^2=1$: $1/PQ^2+1/RS^2$ has fixed value
For $x^2/4+y^2/9=1$: for perpendicular diameters $\frac{1}{PQ^2}+\frac{1}{RS^2}=\frac{1}{4a^2}+\frac{1}{4b^2}=\frac{1}{16}+\frac{1}{36}$. $50(\frac{1}{16}+\frac{1}{36})=50\cdot\frac{52}{576}\approx 4.5$. GIF $=4$.
Correct Answer: 1