Question

Find the equation of the ellipse whose focus is (1, −2), the directrix 3x − 2y + 5 = 0 and eccentricity equal to 1/2.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{S}(1,-2)\mathrm{be}\mathrm{the}\mathrm{focus}\mathrm{and}\mathrm{ZZ}\text{'}\mathrm{be}\mathrm{the}\mathrm{directrix}. \\ \mathrm{Let}P(x,y)\mathrm{be}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{ellipse}\mathrm{and}\mathrm{let}\mathrm{PM}\mathrm{be}\mathrm{the}\mathrm{perpendicular}\mathrm{from}\mathrm{P}\mathrm{on}\mathrm{the}\mathrm{directix}. \\ \mathrm{Then}\mathrm{by}\mathrm{the}\mathrm{definition}\mathrm{of}\mathrm{an}\mathrm{ellipse},\mathrm{we}\mathrm{have}: \\ SP=e.PM,\mathrm{where}e=\frac{1}{2} \\ \Rightarrow S{P}^2=e^2.P{M}^2 \\ \Rightarrow (x-1{)}^2+(y+2{)}^2={\left(\frac{1}{2}\right)}^2.{\left|\frac{3x-2y+5}{\sqrt{(3{)}^2+(-2{)}^2}}\right|}^2 \\ \Rightarrow x^2+1-2x+y^2+4+4y=\left(\frac{1}{4}\right).\left|\frac{9x^2+4y^2+25-12xy-20y+30x}{13}\right| \\ \Rightarrow 52(x^2+1-2x+y^2+4+4y)=9x^2+4y^2+25-12xy-20y+30x \\ \Rightarrow 52x^2+52-104x+52y^2+208+208y=9x^2+4y^2+25-12xy-20y+30x \\ \Rightarrow 43x^2+48y^2-134x+228y+12xy+235=0 \\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{ellipse}. \end{gathered}$$