Ellipse Ques 14
- An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P \frac{1}{2}, 1$. 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
(1996, 2M)
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Answer:
Correct Answer: 14.$\frac{x-\frac{1}{3}^{2}}{1 / 9}+\frac{(y-1)^{2}}{1 / 12}=1$
Solution:
Formula:
- There are two common tangents to the circle $x^{2}+y^{2}=1$ and the hyperbola $x^{2}-y^{2}=1$. These are $x=1$ and $x=-1$. But $x=1$ is nearer to the point $P(1 / 2,1)$.
Therefore, directrix of the required ellipse is $x=1$.
Now, if $Q(x, y)$ is any point on the ellipse, then its distance from the focus is
$ Q P=\sqrt{(x-1 / 2)^{2}+(y-1)^{2}} $
and its distance from the directrix is $|x-1|$.
By definition of ellipse,
$ \begin{aligned} & Q P=e|x-1| \Rightarrow \sqrt {(x-\frac{1}{2})^{2}+(y-1)^{2}}=\frac{1}{2}|x-1| \\ & \Rightarrow \quad (x-\frac{1}{2})^{2}+(y-1)^{2}=\frac{1}{4}(x-1)^{2} \\ & \Rightarrow \quad x^{2}-x+\frac{1}{4}+y^{2}-2 y+1=\frac{1}{4}\left(x^{2}-2 x+1\right) \\ & \Rightarrow \quad 4 x^{2}-4 x+1+4 y^{2}-8 y+4=x^{2}-2 x+1 \\ & \Rightarrow \quad 3 x^{2}-2 x+4 y^{2}-8 y+4=0 \\ & \Rightarrow \quad 3 \quad [(x-\frac{1}{3})^{2}-\frac{1}{9}]+4(y-1)^{2}=0 \\ & \Rightarrow \quad 3 (x-\frac{1}{3})^{2}+4(y-1)^{2}=\frac{1}{3} \\ & \Rightarrow \quad \frac {(x-\frac{1}{3})^{2}}{1 / 9}+\frac{(y-1)^{2}}{1 / 12}=1 \end{aligned} $
BETA
