Question

The equation of the ellipse with focus (−1, 1), directrix x − y + 3 = 0 and eccentricity 1/2 is
(a) 7x² + 2xy + 7y² + 10x + 10y + 7 = 0
(b) 7x² + 2xy + 7y² + 10x − 10y + 7 = 0
(c) 7x² + 2xy + 7y² + 10x − 10y − 7 = 0
(d) none of these

Answer 1

$\left(\mathrm{b}\right)7x^2+7y^2+2xy-10y+10x+7=0$



$$\begin{gathered} \text{Let P(}\mathit{\text{x,y}}\text{) be any point on the ellipse whose focus and}\mathrm{eccentricit}y\mathrm{are}\text{S}\left(-1,1\right)\text{and e=}\frac{1}{2}\text{, respectively.} \\ \text{Let PM be the perpendicular from P on the directrix.} \\ \text{Then}SP=e\times PM \\ \Rightarrow SP=\frac{1}{2}\times PM \\ \Rightarrow 2SP=PM \\ \Rightarrow 4{\left(SP\right)}^2=P{M}^2 \\ \Rightarrow 4\left[{\left(x+1\right)}^2+{\left(y-1\right)}^2\right]={\left(\frac{x-y+3}{\sqrt{1^2+{\left(-1\right)}^2}}\right)}^2 \\ \Rightarrow 4\left[x^2+1+2x+y^2+1-2y\right]=\frac{x^2+y^2+9-2xy-6y+6x}{2} \\ \Rightarrow 8x^2+8+16x+8y^2+8-16y=x^2+y^2+9-2xy-6y+6x \\ \therefore 7x^2+7y^2+2xy-10y+10x+7=0 \\ \text{This is the required equation of the ellipse.} \end{gathered}$$