Question

The auxiliary circle of a family of ellipses passes through origin and makes intercepts of $8$ and $6$ units on the $x-axis$ and the $y-axis$, respectively. If eccentricity of all such family of ellipses is $\frac{1}{2}$, then locus of the focus will be:

• A. $\frac{x^2}{16}-\frac{y^2}{9}=25$
• B. $4x^2+4y^2-32x-24y+75=0$
• C. $\frac{x^2}{16}+\frac{y^2}{9}=25$
• D. $Noneofthese$

Answer 1

Answer: (B)

$4x^2+4y^2-32x-24y+75=0$

Centre of the auxiliary circle $=(4,3)=$ Centre of the elipse

Now, radius of the auxiliary circle is,

$\Rightarrow \sqrt{(4-0{)}^2+(3-0{)}^2}=5$

Ecentricity of the ellipse, $e=\frac{1}{2}$ (Given)

Let focus of the ellipse be $(h,k)$.

Distance of focus from the centre $=ae$

Here,

$a=$ Semi-major axis $=$ radius of the auxiliary circle $=5$

Therefore, the required locus is,

$\sqrt{(h-4{)}^2+(k-3{)}^2}=ae$

$(h-4{)}^2+(k-3{)}^2={\left(\frac{5}{2}\right)}^2$

$h^2+16-8h+k^2+9-6k=\frac{25}{4}$

$4h^2+4k^2-32h-24k+75=0$

Put $h=x,k=y$.

$\Rightarrow 4x^2+4y^2-32x-24y+75=0$

Hence, this is the required equation.