Question Number 76862 by vishalbhardwaj last updated on 31/Dec/19
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{parabola}\: \\ $$$$\mathrm{whose}\:\mathrm{focus}\:\left(−\mathrm{1},−\mathrm{2}\right)\:\mathrm{and}\:\mathrm{directrix} \\ $$$$\mathrm{x}−\mathrm{2y}+\mathrm{3}=\mathrm{0}\:?? \\ $$
Answered by mr W last updated on 31/Dec/19
$${point}\:{P}\left({x},{y}\right)\:{on}\:{parabola} \\ $$$$\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}+\mathrm{2}\right)^{\mathrm{2}} }=\frac{{x}−\mathrm{2}{y}+\mathrm{3}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} }} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}+\mathrm{2}\right)^{\mathrm{2}} =\frac{\left({x}−\mathrm{2}{y}+\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{5}} \\ $$$$\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)+\mathrm{5}\left({y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{4}\right)={x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +\mathrm{9}+\mathrm{2}\left(−\mathrm{2}{xy}+\mathrm{3}{x}−\mathrm{6}{y}\right) \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}{xy}+\mathrm{4}{x}+\mathrm{32}{y}+\mathrm{16}=\mathrm{0} \\ $$
Commented by mr W last updated on 31/Dec/19
