Let and be two ellipses whose centers are at the origin. The major axes of and lie along the -axis and the -axis, respectively. Let be the circle The straight line touches the curves , and at , and respectively. Suppose that If and are the e

Question

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 $P Q = P R = \frac{2 \sqrt{2}}{3} .$ If $e_{1}$ and $e_{2}$ are the eccentricities of $E_{1}$ and $E_{2}$ , respectively, then the correct expression(s) is (are) [2015]

Smart Achievers · Accepted Answer

(1, 2)

$Let E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a > b \dots (i)$

$and E_{2} : \frac{x^{2}}{c^{2}} + \frac{y^{2}}{d^{2}} = 1, where c < d \dots (i i)$

$Also S : x^{2} + {(y - 1)}^{2} = 2 \dots (i i i)$

$Tangent at P (x_{1}, y_{1}) to (i i i) is x + y = 3 \dots (i v)$

$On solving (i i i) and (i v), we get the point of contact P (1, 2)$

$Now, equation of tangent in parametric form,$

$\frac{x - 1}{\frac{- 1}{\sqrt{2}}} = \frac{y - 2}{\frac{1}{\sqrt{2}}} = \pm \frac{2 \sqrt{2}}{3}$ $\Rightarrow x = \frac{1}{3} or \frac{5}{3}$ and $y = \frac{8}{3} or \frac{4}{3}$

$∴$ $Q (\frac{5}{3}, \frac{4}{3}) and R (\frac{1}{3}, \frac{8}{3})$

$Now, equation of tangent to E_{1} at Q is$

$\frac{5 x}{3 a^{2}} + \frac{4 y}{3 b^{2}} = 1$ $which is identical to \frac{x}{3} + \frac{y}{3} = 1$

$\Rightarrow a^{2} = 5 and b^{2} = 4 \Rightarrow e_{1}^{2} = 1 - \frac{4}{5} = \frac{1}{5}$

$And equation of tangent to E_{2} at R is$

$\frac{x}{3 c^{2}} + \frac{8 y}{3 d^{2}} = 1, which is identical to \frac{x}{3} + \frac{y}{3} = 1$

$\Rightarrow c^{2} = 1, d^{2} = 8 \Rightarrow e_{2}^{2} = 1 - \frac{1}{8} = \frac{7}{8}$

$∴$ $e_{1}^{2} + e_{2}^{2} = \frac{43}{40}, e_{1} e_{2} = \frac{\sqrt{7}}{2 \sqrt{10}},$ $| e_{1}^{2} - e_{2}^{2} |$ $= \frac{27}{40}$

Solution

1 $e_{1}^{2} + e_{2}^{2} = \frac{43}{40}$

2 $e_{1} e_{2} = \frac{\sqrt{7}}{2 \sqrt{10}}$

3 $| e_{1}^{2} - e_{2}^{2} |$ $= \frac{5}{8}$

4 $e_{1} e_{2} = \frac{\sqrt{3}}{4}$

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Solution

1 e12+e22=4340

2 e1e2=7210

3 |e12-e22|=58

4 e1e2=34

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1 $e_{1}^{2} + e_{2}^{2} = \frac{43}{40}$

2 $e_{1} e_{2} = \frac{\sqrt{7}}{2 \sqrt{10}}$

3 $| e_{1}^{2} - e_{2}^{2} |$ $= \frac{5}{8}$

4 $e_{1} e_{2} = \frac{\sqrt{3}}{4}$