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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

x^2+y^2=1

intersects the ellipse

x^2+2y^2=4

P

and

Q

, then the locus of the point of intersection of the tangents at

P

and

Q

is:

Aa circle of radius

2

units

Ba parabola with focus

(2,3)

Can ellipse with eccentricity

\dfrac{\sqrt3}{4}

Dan ellipse with length of latus rectum

2

unitscorrect

Solution

Step 1: Write the ellipse as

\dfrac{x^2}{4}+\dfrac{y^2}{2}=1

. Let

R(x_1,y_1)

be the intersection of the tangents at

P,Q

. Then

PQ

is the chord of contact of

R

\frac{xx_1}{4}+\frac{yy_1}{2}=1 \;\Rightarrow\; xx_1+2yy_1-4=0.

Step 2: This chord is tangent to

x^2+y^2=1

, so its distance from the origin is

1

\frac{|-4|}{\sqrt{x_1^2+4y_1^2}}=1 \;\Rightarrow\; x_1^2+4y_1^2=16.

Step 3: The locus is

\dfrac{x^2}{16}+\dfrac{y^2}{4}=1

, an ellipse with

a^2=16,\ b^2=4

. Its latus rectum is

\frac{2b^2}{a}=\frac{2\cdot4}{4}=2.

(Its eccentricity is

\sqrt{1-4/16}=\dfrac{\sqrt3}{2}

, so option (3) is incorrect.) Correct answer: (4)

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