Question

An ellipse has eccentricity $\frac{1}{2}$ and a focus at the point $P(\frac{1}{2},1)$. One of its directrices 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 is:

• A. $3x^2+4y^2-6x-8y+4=0$
• B. $3x^2+4y^2-2x-8y+4=0$
• C. $4x^2+3y^2-8x-6y+4=0$
• D. $4x^2+3y^2-8x-2y+4=0$

Answer 1

The correct option is B $3x^2+4y^2-2x-8y+4=0$

There are two common tangents to the circle $x^2+y^2=1$ and the hyperbola $x^2-y^2=1.$ These are $x=\pm 1$
But $x=1$ is nearer to the point $(\frac{1}{2},1)$
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
$QP=\sqrt{{(x-\frac{1}{2})}^2+{(y-1)}^2}$
And its distance form the directrix is $|x-1|$
By definition of ellipse, $QP=e|x-1|$
$\Rightarrow \sqrt{{(x-\frac{1}{2})}^2+{(y-2)}^2}=\frac{1}{2}|x-1|$
$\Rightarrow {(x-\frac{1}{2})}^2+{(y-2)}^2=\frac{1}{4}{(x-2)}^2$
$\Rightarrow x^2-x+\frac{1}{4}+y^2-2y+1=\frac{1}{4}(x^2-2x+1)$
$\Rightarrow {4x}^2-4x+1+{4y}^2-8y+4=x^2-2x+1$
$\Rightarrow {3x}^2-2x+{4y}^2-8y+4=0$