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If x/a – (y/b)tanθ = 1 and (x/a)tanθ + y/b = 1, then the value of x2/a2 + y2/b2 is
- a)2sec2θ
- b)sec2θ
- c)cos2θ
- d)2cos2θ
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If x/a – (y/b)tanθ = 1 and (x/a)tanθ + y/b = 1, then...
x/a – y/btanθ = 1 and x/atanθ + y/b = 1
Squaring both sides and then adding
⇒ (x/a – y/btanθ)2 + (x/atanθ + y/b)2 = 12 + 12
⇒ (x/a)2 + (y/b)2tan2θ - (2xytanθ) / (ab) + (x/a)2tan2θ + (y/b)2 + (2xytanθ) / (ab) = 2
⇒ x2/a2(1 + tan2θ) + y2/b2(1 + tan2θ) = 2
⇒ (1 + tan2θ) (x2/a2 + y2/b2) = 2
⇒ sec2θ(x2/a2 + y2/b2) = 2 [∵ 1 + tan2θ = sec2θ]
⇒ x2/a2 + y2/b2 = 2/sec2θ
∴ x2/a2 + y2/b2 = 2cos2θ [∵ secθ = 1/cosθ]
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If x/a – (y/b)tanθ = 1 and (x/a)tanθ + y/b = 1, then...
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