JEE Advanced 2015 · Paper 2 · Mathematics Q51

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MathematicsMultiple correct+4 / -2Question 51 of 60

Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$ -axis and the $y$ -axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$ . The straight line $x+y=3$ touches the curves $S$ , $E_1$ and $E_2$ at $P$ , $Q$ and $R$ , respectively. Suppose that $PQ=PR=\dfrac{2\sqrt2}3$ . If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$ , respectively, then the correct expression(s) is(are)

e_1^2+e_2^2=\dfrac{43}{40}

e_1e_2=\dfrac{\sqrt7}{2\sqrt{10}}

|e_1^2-e_2^2|=\dfrac58

e_1e_2=\dfrac{\sqrt3}4

Answer and worked solution

Official answer: A, B

Point P. The circle has centre (0, 1) and radius $\sqrt2$ , so it touches $x+y=3$ at $P=(1,2)$ .

Q and R. They lie on the line at distance $\frac{2\sqrt2}3$ from P: $\left(\frac53,\frac43\right)$ and $\left(\frac13,\frac83\right)$ .

Tangency condition. The tangent to $\frac{x^2}{A}+\frac{y^2}{B}=1$ at $(x_0,y_0)$ is $x+y=3$ only if $\frac{x_0}A=\frac{y_0}B=\frac13$ , i.e. $A=3x_0$ and $B=3y_0$ .

E₁ (major axis along $x$ , so $x_0>y_0$ ): Q $=\left(\frac53,\frac43\right)$ . Then $A=5$ , $B=4$ and $e_1^2=\frac15$ .
E₂ (major axis along $y$ ): R $=\left(\frac13,\frac83\right)$ . Then $A=1$ , $B=8$ and $e_2^2=\frac78$ .

Check the options.

(A) $e_1^2+e_2^2=\frac{43}{40}$ . True.
(B) $e_1e_2=\sqrt{\frac7{40}}=\frac{\sqrt7}{2\sqrt{10}}$ . True.
(C), (D) False.

(A), (B)

Topics:Ellipse Tangents and normals to conics Tangents and normals

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JEE Advanced 2015 · Paper 2, Mathematics question 51