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Maths-

General

Easy

Question

If a tangent to ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction blank$ + $fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction$ = 1 makes an angle $alpha$ with x- axis, then square of length of intercept of tangent cut between axes is-

$a^{2} {t a n}^{2} α + b^{2} {c o t}^{2} α$
$a^{2} {s e c}^{2} α + b^{2} {c o s e c}^{2} α$
$a^{2} {c o s e c}^{2} α + b^{2} {s e c}^{2} α$
None of these

The correct answer is: $a to the power of 2 end exponent s e c to the power of 2 end exponent invisible function application alpha plus b to the power of 2 end exponent c o s e c to the power of 2 end exponent invisible function application alpha$

Let tangent is $fraction numerator x cos invisible function application theta over denominator a end fraction$ + $fraction numerator y sin invisible function application theta over denominator b end fraction$ = 1
Slope is tan $alpha$ = – $fraction numerator b over denominator a end fraction$ $c o t invisible function application theta$ S… (1)
$therefore$ square of length of intercept is
= $left parenthesis a s e c invisible function application theta right parenthesis to the power of 2 end exponent plus left parenthesis b c o s e c invisible function application theta right parenthesis to the power of 2 end exponent$
$therefore$ Length is $L equals a to the power of 2 end exponent plus b to the power of 2 end exponent plus a to the power of 2 end exponent t a n to the power of 2 end exponent invisible function application theta plus b to the power of 2 end exponent c o t to the power of 2 end exponent invisible function application theta$ … (2)
Now use value of tan  from (1)

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