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  • The locus of all such points, so that sum of its distances from <img alt="(-1,2)\; \text{and}\; (-1,7)" src="https://entrancecorner.oncodecogs.com/gif.latex?%28-1%2C2%29%5C%3B%20%5Ctext%7

The locus of all such points, so that sum of its distances from (-1,2)\; \text{and}\; (-1,7) is always 13 , is

Option: 1

\frac{(x+1)^2}{36}+\frac{(2 y-9)^2}{169}=1


Option: 2

\frac{4(x+1)^2}{169}+\frac{\left(y-\frac{9}{2}\right)^2}{36}=1


Option: 3

\frac{(x+1)^2}{36}+\frac{\left(y-\frac{9}{2}\right)^2}{169}=1


Option: 4

None of these


Answers (1)

best_answer

(a) Since, segment joining (-1,2) \: \text{and} \: (-1,7) is parallel to $y$-axis, therefore ellipse is vertical.

\begin{aligned} & 2 b e=5,2 b=13 \\ \\& \Rightarrow \quad e=\frac{5}{13} \\ \\& \therefore \quad a^2=b^2\left(1-e^2\right)=36 \\ & \end{aligned}

and centre is \left(-1, \frac{9}{2}\right). Therefore, equation of the required ellipse is 

\begin{aligned} & \frac{(x+1)^2}{36}+\frac{4\left(y-\frac{9}{2}\right)^2}{169}=1 \\ \\& i.e,\: \: \frac{(x+1)^2}{36}+\frac{(2 y-9)^2}{169}=1 \end{aligned}

Posted by

Devendra Khairwa

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