IIT-JEE 2010 · Paper 2 · Mathematics Q34

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MathematicsSingle correct+3 / -1Question 34 of 57

Passage

Paragraph for Questions 34 to 36

Tangents are drawn from the point $P(3,4)$ to the ellipse $\dfrac{x^2}9+\dfrac{y^2}4=1$ touching the ellipse at points A and B.

The coordinates of A and B are

(3,0)

and

(0,2)

\left(-\dfrac85,\dfrac{2\sqrt{161}}{15}\right)

and

\left(-\dfrac95,\dfrac85\right)

\left(-\dfrac85,\dfrac{2\sqrt{161}}{15}\right)

and

(0,2)

(3,0)

and

\left(-\dfrac95,\dfrac85\right)

Answer and worked solution

Official answer: D

Chord of contact from $P(3,4)$ : $\dfrac{3x}9+\dfrac{4y}4=1$ , i.e. $x=3(1-y)$ .

Intersect with the ellipse.
$(1-y)^2+\frac{y^2}4=1\implies y\left(\frac54y-2\right)=0\implies y=0\text{ or }\frac85.$
The points are $(3,0)$ and $\left(-\frac95,\frac85\right)$ . (Indeed $x=3$ is the tangent at $(3,0)$ through P.)

(D)

Topics:Ellipse Tangents and normals to conics

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IIT-JEE 2010 · Paper 2, Mathematics question 34