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  • If  are the eccentricities of the ellipse, <img alt="\frac{x^{2}}{18}+\frac{y^{2}}{4}=1" src

If e_{1}\: \: and\: \: e_{2} are the eccentricities of the ellipse, \frac{x^{2}}{18}+\frac{y^{2}}{4}=1 and the hyperbola, \frac{x^{2}}{9}-\frac{y^{2}}{4}=1 respectively and \left ( e_{1},e_{2} \right )is a point on the ellipse, 15x^{2}+3y^{2}=k, then k is equal to : 
Option: 1 14
Option: 2 15
Option: 3 17
Option: 4 16
 

Answers (1)

best_answer

 

 

What is Ellipse? -

Ellipse

\mathbf{\frac{\mathbf{x}^{2}}{\mathbf{a}^{2}}+\frac{\mathbf{y}^{2}}{\mathbf{b}^{2}}=1} \quad \text { where }, \mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)_{(\mathrm{a}>\mathrm{b})}

 

  1. a > b 

  2.  the length of the major axis is 2a 

  3.  the coordinates of the vertices are (±a, 0) 

  4.  the length of the minor axis is 2b 

  5.  the coordinates of the co-vertices are (0, ±b)

 

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What is Hyperbola? -

Hyperbola:

Eccentricity of Hyperbola: 

\\\mathrm{Equation\;of\;the\;hyperbola\;is\;\;\;\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}\\\text{we have,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;b^2=a^2\left ( e^2-1 \right )}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e^2=\frac{b^2+a^2}{a^2}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{b^2}{a^2} \right )}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{2b}{2a} \right )^2}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e=\sqrt{1+\left ( \frac{conjugate \;axis}{transverse \;axis} \right )^2}}

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\begin{aligned} &e_{1}=\sqrt{1-\frac{4}{18}}=\sqrt{\frac{7}{9}}=\frac{\sqrt{7}}{3}\\ &\mathrm{e}_{2}=\sqrt{1+\frac{4}{9}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}\\ &15 e_{1}^{2}+3 e_{2}^{2}=k \Rightarrow \quad k=15\left(\frac{7}{9}\right)+3\left(\frac{13}{9}\right) \end{aligned}

So, k= 16

Posted by

avinash.dongre

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