Question Number 51612 by peter frank last updated on 29/Dec/18
$${Find}\:{the}\:{equation}\:{of} \\ $$$${the}\:{ellipse}\:{with}\:{ecentricity} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{the}\:{focus}\:\left(\mathrm{2},\mathrm{1}\right) \\ $$$${Does}\:{the}\:{line}\:{x}=\mathrm{3}\:{touches} \\ $$$${ellipse}.{if}\:{so}\:{at}\:{what}\: \\ $$$${point}?{if}\:{line}\:{x}=\mathrm{5}\:{is}\:{the} \\ $$$${line}\:{of}\:{direction}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18
Answered by peter frank last updated on 29/Dec/18
$$\frac{{PS}}{{PM}}={e} \\ $$$${e}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${P}=\left({x},{y}\right) \\ $$$${S}=\left(\mathrm{2},\mathrm{1}\right) \\ $$$${M}=\left(\mathrm{5},{y}\right)\Rightarrow{directrix} \\ $$$${PS}=\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${PM}=\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−{y}\right)} \\ $$$$\frac{{PS}}{{PM}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{8}{y}−\mathrm{5}=\mathrm{0} \\ $$$${x}=\mathrm{3} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$$${from} \\ $$$${b}^{\mathrm{2}} =\mathrm{4}{ac} \\ $$$$\left(−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${hence}\:\:\:{x}=\mathrm{3}\:{touches}\:{the} \\ $$$${ellipse} \\ $$$$ \\ $$