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Max Distance: Centre to Chord of Contact | JEE

JEE Maths question with a full step-by-step solution.

Question
The maximum distance of the centre of the ellipse x216+y29=1\dfrac{x^2}{16}+\dfrac{y^2}{9}=1 from the chord of contact of mutually perpendicular tangents is:
A125\dfrac{12}{5}
B45\dfrac{4}{5}
C325\dfrac{32}{5}
D165\dfrac{16}{5}correct
Solution
Step 1: Mutually perpendicular tangents meet on the director circle x2+y2=a2+b2=16+9=25x^2+y^2=a^2+b^2=16+9=25, of radius 55. Take a point (5cos⁡θ,5sin⁡θ)(5\cos\theta,5\sin\theta) on it. Step 2: The chord of contact from this point is x⋅5cos⁡θ16+y⋅5sin⁡θ9=1\dfrac{x\cdot5\cos\theta}{16}+\dfrac{y\cdot5\sin\theta}{9}=1, i.e.
45cos⁡θ x+80sin⁡θ y−144=0.45\cos\theta\,x+80\sin\theta\,y-144=0.
 Step 3: Distance from the centre (0,0)(0,0) to this chord is
d=144(45cos⁡θ)2+(80sin⁡θ)2=144581cos⁡2θ+256sin⁡2θ=144581+175sin⁡2θ.d=\frac{144}{\sqrt{(45\cos\theta)^2+(80\sin\theta)^2}}=\frac{144}{5\sqrt{81\cos^2\theta+256\sin^2\theta}}=\frac{144}{5\sqrt{81+175\sin^2\theta}}.
 Step 4: dd is greatest when sin⁡2θ=0\sin^2\theta=0, giving
dmax⁡=1445⋅9=165.d_{\max}=\frac{144}{5\cdot9}=\frac{16}{5}.
 Correct answer: (4)
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