Question Number 31846 by momo last updated on 15/Mar/18
$$\mathrm{25}\left[\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} \right]=\left(\mathrm{3}{x}−\mathrm{4}{y}+\mathrm{7}\right)^{\mathrm{2}} \\ $$$${is}\:{the}\:{equation}\:{of}\:{parabola}.{Find} \\ $$$${length}\:{of}\:{latus}\:{rectum} \\ $$
Commented by momo last updated on 16/Mar/18
$${how} \\ $$
Commented by momo last updated on 16/Mar/18
$${explain}\:{with}\:{sol}^{{n}} \\ $$
Answered by Tinkutara last updated on 16/Mar/18
$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}{x}−\mathrm{4}{y}+\mathrm{7}}{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} }=\mid\frac{\mathrm{3}{x}−\mathrm{4}{y}+\mathrm{7}}{\mathrm{5}}\mid \\ $$$${Focus}=\left(\mathrm{2},\mathrm{3}\right) \\ $$$${Directrix}\:{is}\:\mathrm{3}{x}−\mathrm{4}{y}+\mathrm{7}=\mathrm{0} \\ $$$${Distance}\:{of}\:{focus}\:{from}\:{directrix}=\mathrm{2}{a} \\ $$$$\mathrm{2}{a}=\mid\frac{\mathrm{6}−\mathrm{12}+\mathrm{7}}{\mathrm{5}}\mid=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{4}{a}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$
Commented by momo last updated on 16/Mar/18
$${thanks}\:{sir} \\ $$