Question

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

• A. $4x^2+4y^2+32x-24y+75=0$
• B. $4x^2+4y^2-32x-24y+75=0$
• C. $4x^2+4y^2-32x+24y+75=0$
• D. $4x^2+4y^2-32x-24y-75=0$

Answer 1

The correct option is B $4x^2+4y^2-32x-24y+75=0$



Centre of family of ellipse is $(\frac{0+8}{2},\frac{6+0}{2})=(4,3)$

Length of major axis $2a=\sqrt{8^2+6^2}$
$\Rightarrow a=5$

Distance of focus from centre $=ae$
$=5\times \frac{1}{2}=\frac{5}{2}$

If $(h,k)$ is the coordinates of focus, then
$(h-4{)}^2+(k-3{)}^2=\frac{25}{4}$
Hence, required locus is
$\Rightarrow 4x^2+4y^2-32x-24y+75=0$