• question_answer 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       [IIT 1996] A)            $\frac{{{(x-1/3)}^{2}}}{1/9}+\frac{{{(y-1)}^{2}}}{1/12}=1$ B)            $\frac{{{(x-1/3)}^{2}}}{1/9}+\frac{{{(y+1)}^{2}}}{1/12}=1$ C)            $\frac{{{(x-1/3)}^{2}}}{1/9}-\frac{{{(y-1)}^{2}}}{1/12}=1$ D)            $\frac{{{(x-1/3)}^{2}}}{1/9}-\frac{{{(y+1)}^{2}}}{1/12}=1$

Correct Answer: A

Solution :

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 $x=-1$                   Out of these, $x=1$ is nearer to the point $P(1/2,1)$.                   Thus a directrix of the required ellipse is $x=1.$            If $Q(x,y)$ is any point on the ellipse, then its distance from the focus is $QP=\sqrt{{{\left( x-\frac{1}{2} \right)}^{2}}+{{(y-1)}^{2}}}$ and its distance from the directrix $x=1$is $|x-1|$.            By definition of ellipse, $QP=e|x-1|$            $\Rightarrow \sqrt{{{\left( x-\frac{1}{2} \right)}^{2}}+{{(y-1)}^{2}}}=\frac{1}{2}|x-1|$                    Þ$3{{x}^{2}}-2x+4{{y}^{2}}-8y+4=0$or$\frac{{{\left( x-\frac{1}{3} \right)}^{2}}}{1/9}+\frac{{{(y-1)}^{2}}}{1/12}=1$.

You need to login to perform this action.
You will be redirected in 3 sec