Question

The combined equation of directrices of the ellipse whose eccentricity is $\frac{1}{2}$, one of the foci is $(1,1)$ and minor axis is $x+y=4$ is

• A. $x^2+y^2+2xy+8(x+y)+48=0$
• B. $x^2+y^2+2xy-8(x+y)+48=0$
• C. $x^2+y^2+2xy-8(x+y)-48=0$
• D. $x^2+y^2+2xy+8(x+y)-48=0$

Answer 1

Answer: (C)

Equation of major axis: $x-y=\lambda$, passes through $(1,1)\Rightarrow \lambda =0$ $3/$ ellisC $\therefore x-y=0$
Hence, centre is $(2,2)$
Now, $CS=$ ae
$\Rightarrow \sqrt{2}=a\cdot \frac{1}{2}\Rightarrow a=2\sqrt{2}\text{. }$
Equation of directrix: $x+y=\lambda$
$\text{ perpendicular from centre to directrix }=\frac{a}{e}$
$\Rightarrow \left|\frac{\lambda -4}{\sqrt{2}}\right|=\frac{2\sqrt{2}}{1/2}=4\sqrt{2}$
$\lambda -4=\pm 8$
$\lambda =12\text{ or }-4$
$\therefore \text{ equation of directrix are }x+y-12=0\text{ and }x+y+4=0$
$\therefore \text{ combined equation: }(x+y-12)(x+y+4)=0$
$(x+y{)}^2-8(x+y)-48=0$