Question

Equation of the directrices of the ellipse $4x^2+y^2-8x+2y+1=0$ are:

• A. $\sqrt{3}y+\sqrt{3}\pm 5=0$
• B. $\sqrt{3}y+\sqrt{3}\pm 4=0$
• C. $\sqrt{3}y+\sqrt{3}\pm 7=0$
• D. $\sqrt{3}y+\sqrt{3}\pm 8=0$

Answer 1

The correct option is B $\sqrt{3}y+\sqrt{3}\pm 4=0$

The equation of the ellipse is : $4x^2+y^2-8x+2y+1=0$

$\to$ $\left[\right(2x{)}^2-8x+4]+[y^2+2y+1]-4=0$

$\to$ $4(x-1{)}^2+(y+1{)}^2=4$

$\to$ $\frac{(x-1{)}^2}{1^2}+\frac{(y+1{)}^2}{2^2}=1$

If the equation of the ellipse is $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$ where $(b>a)$ then the equation of the directrices are given by

$y-k=\pm \frac{b}{e}$ where $e=\frac{\sqrt{b^2-a^2}}{a}$ is the eccentricity of the ellipse.

Here $k=-1$ and$e=\frac{\sqrt{2^2-1^2}}{1}=\frac{\sqrt{3}}{2}$

Here the directrices are given by :

$y-(-1)=\pm \frac{2}{\left(\frac{\sqrt{3}}{2}\right)}$

$\to$ $\sqrt{3}(y+1)=\pm 4$

$\to$ $\sqrt{3}y+\sqrt{3}\pm 4=0$