Question

The equation of directrix of a conic is
$L:x+y-1=0$ and the focus is the point $(0,0)$. Find the equation of the conic if its eccentricity is
$\frac{1}{\sqrt{2}}$

• A. $3x^2+3y^2+2xy+2x+2y-1=0$
• B. $3x^2+3y^2+2xy+2x+2y+1=0$
• C. $3x^2+3y^2+2xy-2x-2y+1=0$
• D. $3x^2-3y^2+2xy+2x-2y+1=0$

Answer 1

The correct option is C

$3x^2+3y^2+2xy-2x-2y+1=0$

we will use the definition of conic to arrive at its equation.

$\frac{PS}{PM}=e$

P is our moving point and let its coordinate be (h,k)

PS = $\sqrt{(h-0{)}^2+(k-0{)}^2}$ = $\sqrt{h^2+k^2}$

PM = perpendicular distance of P from L

=$\left|\begin{array}{c}\frac{h+k-1}{\sqrt{2}}\end{array}\right|$

e = $\frac{1}{\sqrt{2}}$ (given)

$\frac{PS}{PM}=e\Rightarrow \sqrt{h^2+k^2}=\frac{1}{\sqrt{2}}\times \left|\begin{array}{c}\frac{h+k-1}{\sqrt{2}}\end{array}\right|$

$\Rightarrow h^2+k^2=\frac{(h+k-1{)}^2}{2^2}$

$\Rightarrow 4h^2+4k^2=h^2+k^2-2hk+1-2h-2k$

$\Rightarrow 3h^2+3k^2-2hk-2h-2k+1=0$
we will replace (h,k) with (x,y) to get the equation of conic
$\Rightarrow 3x^2+3y^2-2xy-2x-2y+1=0$