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JEE MainMathematicsToughConic Sections

Let Y_1 and Y_2 be the foot of the perpendiculars on the
auxiliary circle from the foci upon any tangent at P on the ellipse. The
tangent at Y_1 and Y_2 to the auxiliary circle intersect at Q. Then
which of the following holds:
ekuk nh?kZo`Ùk

Show answer and solution
Let the given ellipse
bex2a2+y2b2=1(a>b)\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b) and auxiliary
circle of this ellipse is x^2 + y^2 = a^2
Let the tangents to auxiliary circle at points Y_1& Y_2 intersect at
Q(h,k), then Y_1Y_2 will be the chord of contact of Q w.r.t.
auxiliary circle.
Chord of contact w.r.t. the circle x^2 + y^2 = a^2 is
xh + yk = a^2....(i)
This must be same as tangent at P(θ)on
ellipsex2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1
i.e. xacosθ+ybsinθ\frac{x}{a}cos\theta + \frac{y}{b}sin\theta = 1.....(ii)
Comparing (i) and (ii)
h = a cos θ, k = a2sinθb\frac{a^{2}sin\theta}{b}
∴\therefore locus of Q:
x2a2+y2(a2b)2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{\left( \frac{a^{2}}{b} \right)^{2}}
= 1
ekuk fd nh?kZo`Ùk
x2a2+y2b2=1(a>b)\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b) gS rFkk bl
nh?kZo`Ùk o`Ùk x^2 + y^2 = a^2 gSA
ekuk fd lgk;d o`Ùk ds fcUnq Y_1
ij feyrh gSA rc Y_1Y_2, Q ds
o`Ùk x^2 + y^2 = a^2 thok gSA
xh + yk = a^2....(i)
nh?kZo`Ùk x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 ds P(θ) ij
i.e. xacosθ+ybsinθ\frac{x}{a}cos\theta + \frac{y}{b}sin\theta = 1.....(ii)
(i) o (ii) dh rqyuk
h=acosθ,k=a2sinθbh = acos\theta,k = \frac{a^{2}sin\theta}{b}
∴\therefore Q

Source: Grass-G / M-27 Mathematics Paper 2 15 Jan.docx

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