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  • An ellipse has an eccentricity  and one focus at the point  <img alt="P\left

An ellipse has an eccentricity \frac{1}{2} and one focus at the point  P\left(\frac{1}{2}, 1\right). It's one directrix is the common tangent, nearer to the point P, to the circle x^2+y^2=1 and hyperbola x^2-y^2=1 The equation of the ellipse in the standard form is ?

Option: 1

\begin{aligned} \frac{\left(x+\frac{1}{3}\right)^2}{\left(\frac{1}{3}\right)^2}-\frac{(y+1)^2}{\left(\frac{1}{2} \sqrt{2}\right)^2} \\ \end{aligned}


Option: 2

\frac{\left(x^2+\frac{2}{9}\right)^2}{\left(\frac{2}{9}\right)^2}-\frac{(y+5)^2}{\left(\frac{5}{7} \sqrt{2}\right)^2} \\


Option: 3

\frac{\left(x-\frac{1}{3}\right)^2}{\left(\frac{1}{3}\right)^2}+\frac{(y-1)^2}{\left(\frac{1}{2} \sqrt{2}\right)^2}


Option: 4

\frac{\left(x^2-\frac{2}{9}\right)^2}{\left(\frac{2}{9}\right)^2}+\frac{(y-5)^2}{\left(\frac{5}{7} \sqrt{2}\right)^2}


Answers (1)

best_answer

Rough graph of  x^2+y^2=1(\text { Circle }) —-----(i)

And  \left.x^2-y^2=1 \text { (hyperbola }\right) —-------(ii)

is shown 

It is clear from the graph that there are two common tangents to the curve (1) and (2) namely x = 1 and x = -1 out of which x = 1 is near to the point P.

Also, e=\frac{1}{2}  , focus (\frac{1}{2},1) , then equation ellipse is given by

\begin{aligned} &\left(x-\frac{1}{2}\right)^2+(y-1)^2=\frac{1}{4}(x-1)^2 \\ & \Rightarrow \frac{\left(x-\frac{1}{3}\right)^2}{\left(\frac{1}{3}\right)^2}+\frac{(y-1)^2}{\left(\frac{1}{2} \sqrt{2}\right)^2} \\ \end{aligned}

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vinayak

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