Let P ( 2 3 7 , 6 7 ) , Q , R and S be four points on ellipse 9 x 2 + 4 y 2 = 36 . Let PQ and RS be mutually perpendicular chords and pass through the centre of ellipse, then the value of [ 50 P Q 2 + 50 R S 2 ] = _ _ _ _ where [ . ] denotes GI

Question

Let $P (\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}), Q, R$ and S be four points on ellipse $9 x^{2} + 4 y^{2} = 36$ . Let PQ and RS be mutually perpendicular chords and pass through the centre of ellipse, then the value of $[\frac{50}{P Q^{2}} + \frac{50}{R S^{2}}] =____$ where $[.]$ denotes GIF

Smart Achievers · Accepted Answer

(1)

Given ellipse:

$9 x^{2} + 4 y^{2} = 36$

and

$P (\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}})$

Since $P Q$ passes through the centre, $P Q$ is a diameter.

$O P^{2} = \frac{12}{7} + \frac{36}{7} = \frac{48}{7}$

$P Q^{2} = 4 O P^{2} = \frac{192}{7}$

Since $R S ⊥ P Q$ , slope of $O P = \sqrt{3}$ , so slope of $O R = - \frac{1}{\sqrt{3}}$ .

Put $y = - \frac{x}{\sqrt{3}}$ in the ellipse:

$9 x^{2} + \frac{4 x^{2}}{3} = 36$

$x^{2} = \frac{108}{31}$

Hence,

$O R^{2} = x^{2} + y^{2} = \frac{144}{31}$

$R S^{2} = 4 O R^{2} = \frac{576}{31}$

Therefore,

$\frac{50}{P Q^{2}} + \frac{50}{R S^{2}} = \frac{350}{192} + \frac{1550}{576} = \frac{325}{72} \approx 4.51$

Thus,

$[\frac{50}{P Q^{2}} + \frac{50}{R S^{2}}] = 4$

Solution

1 4

2 5

3 6

4 7

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Solution

Q. Let P(237,67), Q, R and S be four points on ellipse 9x2+4y2=36. Let PQ and RS be mutually perpendicular chords and pass through the centre of ellipse, then the value of [50PQ2+50RS2]=____ where [.] denotes GIF

1 4

2 5

3 6

4 7

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