An ellipse has eccentricity frac{1}{2} and on | vedclass

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(A) The common tangents to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$ are $x = 1$ and $x = -1$.Out of these, $x = 1$ is nearer to th

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$\frac{(x - 1/3)^2}{1/9} + \frac{(y - 1)^2}{1/12} = 1$

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(A) The common tangents to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$ are $x = 1$ and $x = -1$.Out of these, $x = 1$ is nearer to the point $P\left( \frac{1}{2}, 1 ight)$.Thus, the directrix of the required ellipse is $x = 1$.Let $Q(x, y)$ be any point on the ellipse.By the definition of an ellipse, the distance from the focus $P$ is $QP = e 	imes (	ext{distance from directrix})$.$QP = \sqrt{(x - 1/2)^2 + (y - 1)^2}$ and the distance from the directrix $x = 1$ is $|x - 1|$.Given $e = 1/2$, we have $\sqrt{(x - 1/2)^2 + (y - 1)^2} = \frac{1}{2} |x - 1|$.Squaring both sides: $(x - 1/2)^2 + (y - 1)^2 = \frac{1}{4}(x - 1)^2$.$x^2 - x + 1/4 + (y - 1)^2 = \frac{1}{4}(x^2 - 2x + 1)$.$x^2 - x + 1/4 + (y - 1)^2 = \frac{1}{4}x^2 - \frac{1}{2}x + 1/4$.$\frac{3}{4}x^2 - \frac{1}{2}x + (y - 1)^2 = 0$.Multiply by $4/3$: $x^2 - \frac{2}{3}x + \frac{4}{3}(y - 1)^2 = 0$.$(x^2 - \frac{2}{3}x + 1/9) + \frac{4}{3}(y - 1)^2 = 1/9$.$(x - 1/3)^2 + \frac{4}{3}(y - 1)^2 = 1/9$.Dividing by $1/9$: $\frac{(x - 1/3)^2}{1/9} + \frac{(y - 1)^2}{1/12} = 1$.

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

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