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Question: If any tangent to the ellipse \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\)intercepts equal leng...

If any tangent to the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1intercepts equal lengths l on the axes, then l=

A

a2+b2a^{2} + b^{2}

B

a2+b2\sqrt{a^{2} + b^{2}}

C

(a2+b2)2(a^{2} + b^{2})^{2}

D

None of these

Answer

a2+b2\sqrt{a^{2} + b^{2}}

Explanation

Solution

The equation of any tangent to the given ellipse is

xacos⁡θ+ybsin⁡θ=1\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1

This line meets the coordinate axes at P(acos⁡θ,0)P\left( \frac{a}{\cos\theta},0 \right)and Q(0,bsin⁡θ)Q\left( 0,\frac{b}{\sin\theta} \right)

∴acos⁡θ=l=bsin⁡θ\frac{a}{\cos\theta} = l = \frac{b}{\sin\theta}⇒ cos⁡θ=al\cos\theta = \frac{a}{l}and sin⁡θ=bl\sin\theta = \frac{b}{l} ⇒

cos⁡2θ+sin⁡2θ=a2l2+b2l2\cos^{2}\theta + \sin^{2}\theta = \frac{a^{2}}{l^{2}} + \frac{b^{2}}{l^{2}}⇒l2=a2+b2l^{2} = a^{2} + b^{2}⇒l=a2+b2l = \sqrt{a^{2} + b^{2}}