P SP1 and QS Q1 are two perpendicular focal chords of x216−y29=1, then 1PP1+1QQ1=______

consider lr=1+ecos⁡θ⇒e=16+916=54;l=b2a=94⇒lSP=1+ecos⁡θ⇒SP=l1+ecos⁡θ⇒lSP1=1−ecos⁡θ⇒SP1=l1−ecos⁡θ⇒lSQ=1−esin⁡θ⇒SQ=l1−esin⁡θ⇒lSQ1=1+esin⁡θ⇒SQ1=l1+ecos⁡θ⇒PP1=SP+SP1=2l1−e2cos2⁡θQQ1=2l1−e2sin2⁡θ∴1PP1+1QQ1=2−e22l=2−5422⋅94=7×216×9=772=63648

P SP1 and QS Q1 are two perpendicular focal chords of x216−y29=1, then 1PP1+1QQ1=______

consider lr=1+ecos⁡θ⇒e=16+916=54;l=b2a=94⇒lSP=1+ecos⁡θ⇒SP=l1+ecos⁡θ⇒lSP1=1−ecos⁡θ⇒SP1=l1−ecos⁡θ⇒lSQ=1−esin⁡θ⇒SQ=l1−esin⁡θ⇒lSQ1=1+esin⁡θ⇒SQ1=l1+ecos⁡θ⇒PP1=SP+SP1=2l1−e2cos2⁡θQQ1=2l1−e2sin2⁡θ∴1PP1+1QQ1=2−e22l=2−5422⋅94=7×216×9=772=63648

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$P {SP}^{1} and QS Q^{1}$ are two perpendicular focal chords of $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ , then $\frac{1}{{PP}^{1}} + \frac{1}{{QQ}^{1}} =______$

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$\frac{63}{648}$

$\frac{63}{128}$

$\frac{63}{256}$

$\frac{63}{32}$

answer is A.

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Detailed Solution

consider $\frac{l}{r} = 1 + ecos θ$

$\begin{array}{l} \Rightarrow e = \sqrt{\frac{16 + 9}{16}} = \frac{5}{4}; l = \frac{b^{2}}{a} = \frac{9}{4} \\ \Rightarrow \frac{l}{SP} = 1 + ecos θ \Rightarrow SP = \frac{l}{1 + ecos θ} \\ \Rightarrow \frac{l}{{SP}^{1}} = 1 - ecos θ \Rightarrow {SP}^{1} = \frac{l}{1 - ecos θ} \\ \Rightarrow \frac{l}{SQ} = 1 - esin θ \Rightarrow SQ = \frac{l}{1 - esin θ} \\ \Rightarrow \frac{l}{{SQ}^{1}} = 1 + esin θ \Rightarrow {SQ}^{1} = \frac{l}{1 + ecos θ} \\ \Rightarrow {PP}^{1} = SP + {SP}^{1} = \frac{2 l}{1 - e^{2} \cos^{2} θ} {QQ}^{1} = \frac{2 l}{1 - e^{2} \sin^{2} θ} \\ ∴ \frac{1}{{PP}^{1}} + \frac{1}{{QQ}^{1}} = \frac{2 - e^{2}}{2 l} = \frac{2 - {(\frac{5}{4})}^{2}}{2 \cdot \frac{9}{4}} = \frac{7 \times 2}{16 \times 9} = \frac{7}{72} = \frac{63}{648} \end{array}$