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Locus of Tangent Intersection of Chord of Contact | JEE

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

Question
If a variable tangent of the circle x2+y2=1x^2+y^2=1 intersects the ellipse x2+2y2=4x^2+2y^2=4 at PP and QQ, then the locus of the point of intersection of the tangents at PP and QQ is:
Aa circle of radius 22 units
Ba parabola with focus (2,3)(2,3)
Can ellipse with eccentricity 34\dfrac{\sqrt3}{4}
Dan ellipse with length of latus rectum 22 unitscorrect
Solution
Step 1: Write the ellipse as x24+y22=1\dfrac{x^2}{4}+\dfrac{y^2}{2}=1. Let R(x1,y1)R(x_1,y_1) be the intersection of the tangents at P,QP,Q. Then PQPQ is the chord of contact of RR:
xx14+yy12=1  ⇒  xx1+2yy1−4=0.\frac{xx_1}{4}+\frac{yy_1}{2}=1 \;\Rightarrow\; xx_1+2yy_1-4=0.
 Step 2: This chord is tangent to x2+y2=1x^2+y^2=1, so its distance from the origin is 11:
∣−4∣x12+4y12=1  ⇒  x12+4y12=16.\frac{|-4|}{\sqrt{x_1^2+4y_1^2}}=1 \;\Rightarrow\; x_1^2+4y_1^2=16.
 Step 3: The locus is x216+y24=1\dfrac{x^2}{16}+\dfrac{y^2}{4}=1, an ellipse with a2=16, b2=4a^2=16,\ b^2=4. Its latus rectum is
2b2a=2⋅44=2.\frac{2b^2}{a}=\frac{2\cdot4}{4}=2.
 (Its eccentricity is 1−4/16=32\sqrt{1-4/16}=\dfrac{\sqrt3}{2}, so option (3) is incorrect.) Correct answer: (4)
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