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Maths-

General

Easy

Question

PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R and
$2 alpha$ , $2 beta comma blank 2 gamma$ respectively then tan $beta$ tan $gamma$ is equal to -

cot $α$
cot² $α$
2 cot $α$
None of these

The correct answer is: cot² $alpha$

$fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1$
P (a cos 2 $alpha$ , b sin 2 $alpha$ ), Q (a cos 2 $beta$ , b sin 2 $beta$ )
R (a cos 2 $gamma$ , b sin 2 $gamma$ )
chord's PQ equation
$fraction numerator x over denominator a end fraction$ cos ( $alpha$ + $beta$ ) + $fraction numerator y over denominator b end fraction$ sin ( $alpha$ + $beta$ ) = cos ( $alpha$ – $beta$ )
PQ passes through the focus (ae, 0)
e = $fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction$
PR passes through the focus (– ae, 0) the
– e = $fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction$
$fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction$ = – $fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction$
Apply componendo and dividendo, we get
$fraction numerator cos invisible function application left parenthesis alpha plus beta right parenthesis plus cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis minus cos invisible function application left parenthesis alpha minus beta right parenthesis end fraction$ = $fraction numerator cos invisible function application left parenthesis alpha plus gamma right parenthesis minus cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis plus cos invisible function application left parenthesis alpha minus gamma right parenthesis end fraction$
$fraction numerator 2 cos invisible function application alpha cos invisible function application beta over denominator 2 sin invisible function application alpha sin invisible function application beta end fraction$ = $fraction numerator 2 sin invisible function application alpha sin invisible function application gamma over denominator 2 cos invisible function application alpha cos invisible function application gamma end fraction$
tan $beta$ tan $gamma$ = cot² $alpha$

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