If PSP 1 is a focal chord of 25 x 2 + 9 y 2 = 1 and SP = 3 then 61 ( S 1 P 1 ) − 56 = Where S , S 1 are foci of ellipse.

∵ SP + S 1 P = 2 a = 10 ⇒ S 1 P = 7 ∵ S 1 P 1 + S 1 P 1 1 = 1 2 = ( a b ) 2 2 ⇒ S 1 P 1 1 = 9 10 − 7 1 = 63 61 ∴ 61 ( S 1 P 1 ) − 56 = 7

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MathematicsEllipseChordsMedium2 minQB

MathematicsMediuminteger

If $PSP^{1}$ is a focal chord of $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ and $SP = 3$ then $61 (S^{1} P^{1}) - 56 =$ Where $S, S^{1}$ are foci of ellipse.

Answer:

Solution:

$∵ SP + S^{1} P = 2 a = 10 \Rightarrow S^{1} P = 7$ $∵ \frac{1}{S ^{1} P} + \frac{1}{S ^{1} P ^{1}} = \frac{2}{1} = \frac{2}{( \frac{b}{a} ) ^{2}} \Rightarrow \frac{1}{S ^{1} P ^{1}} = \frac{10}{9} - \frac{1}{7} = \frac{61}{63}$ $∴ 61 (S^{1} P^{1}) - 56 = 7$

Stream:JEESubject:MathematicsTopic:EllipseSubtopic:Chords

⏱ 2mℹ️ Source: QB

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