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Question

Question: The locus of the mid points of the portion of the tangents to the ellipse intercepted between the ax...

The locus of the mid points of the portion of the tangents to the ellipse intercepted between the axes is

A

a2y2+b2x2=4x2y2a^{2}y^{2} + b^{2}x^{2} = 4x^{2}y^{2}

B

a2x2+b2y2=4x2y2a^{2}x^{2} + b^{2}y^{2} = 4x^{2}y^{2}

C

x2+y2=a2x^{2} + y^{2} = a^{2}

D

x2+y2=b2x^{2} + y^{2} = b^{2}

Answer

a2y2+b2x2=4x2y2a^{2}y^{2} + b^{2}x^{2} = 4x^{2}y^{2}

Explanation

Solution

Equation of the tangent to the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at (α cos θ, b sin θ) is xcosθ2+ysinθ2=1\frac{x\cos\theta}{2} + \frac{y\sin\theta}{2} = 1

∴ Coordinates of the mid point of the portion

intercepted between the axes is

(α2secθ,b2cosecθ)\left( \frac{\alpha}{2}\sec\theta,\frac{b}{2}\cos ec\theta \right) ∴ x = α2secθcosθ=α2x\frac{\alpha}{2}\sec\theta \Rightarrow \cos\theta = \frac{\alpha}{2x}

and y = b2\frac{b}{2}cosec θ ⇒ sinθ = b2y\frac{b}{2y}a24x2+b24y2\frac{a^{2}}{4x^{2}} + \frac{b^{2}}{4y^{2}}=1

Thus, required locus is a2y2+b2x2=4x2y2a^{2}y^{2} + b^{2}x^{2} = 4x^{2}y^{2}