Question Number 88378 by Rio Michael last updated on 10/Apr/20
$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{parabola}\:\mathrm{with}\:\mathrm{focus}\:\left(\mathrm{3},\mathrm{3}\right) \\ $$$$\mathrm{and}\:\mathrm{directrix}\:\:{y}\:=\:\mathrm{0} \\ $$
Commented by john santu last updated on 10/Apr/20
$${o}\:{yes}.\:{should}\:{be}\:{a}\:{vertex}\: \\ $$$${at}\:\left(\mathrm{3},\:\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$
Commented by john santu last updated on 10/Apr/20
Commented by john santu last updated on 10/Apr/20
$${focus}\:\left({a},\:{b}+{p}\right)\:=\:\left(\mathrm{3},\mathrm{3}\right)\:\Rightarrow\begin{cases}{{a}=\mathrm{3}}\\{{b}+{p}\:=\:\mathrm{3}}\end{cases} \\ $$$${directrix}\:{y}\:=\:{b}−{p}\:=\:\mathrm{0}\:,\:{b}={p} \\ $$$${b}\:=\:{p}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${equation}\:{of}\:{a}\:{parabola} \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{4}.\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\left({y}−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{6}\:\left({y}−\frac{\mathrm{3}}{\mathrm{2}}\right)=\:\mathrm{6}{y}−\mathrm{9} \\ $$
Commented by mr W last updated on 10/Apr/20
$${correct} \\ $$
Answered by mr W last updated on 10/Apr/20
$${definition}\:{of}\:{parabola}: \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\left({y}−\mathrm{0}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{6}{y}+\mathrm{18}=\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{6}}−{x}+\mathrm{3} \\ $$
Commented by jagoll last updated on 10/Apr/20
$$\mathrm{if}\:\mathrm{the}\:\mathrm{focus}\:\mathrm{at}\:\left(\mathrm{2},\mathrm{5}\right)\:\mathrm{and}\:\mathrm{dirextrix} \\ $$$$\mathrm{x}+\mathrm{4}\:=\:\mathrm{0},\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{should}\:\mathrm{be}\: \\ $$$$\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{5}\right)^{\mathrm{2}} \:=\:\left(\mathrm{x}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$−\mathrm{4x}+\mathrm{4}+\mathrm{y}^{\mathrm{2}} −\mathrm{10y}+\mathrm{25}=\:\mathrm{8x}+\mathrm{16} \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{10y}+\mathrm{13}\:=\:\mathrm{12x} \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{12x}−\mathrm{10y}+\mathrm{13}\:=\:\mathrm{0}\:?\: \\ $$$$\mathrm{it}\:\mathrm{correct}\:? \\ $$
Commented by mr W last updated on 10/Apr/20
$${yes}. \\ $$$${but}\:{what}\:{if}\:{the}\:{focus}\:{is}\:{at}\:\left(\mathrm{2},\mathrm{5}\right)\:{and}\:{the} \\ $$$${dirextrix}\:{is}\:{y}+\mathrm{2}{x}+\mathrm{1}=\mathrm{0}\:? \\ $$
Commented by jagoll last updated on 10/Apr/20
$$\mathrm{does}\:\mathrm{this}\:\mathrm{matter}\:\mathrm{wrong}\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 10/Apr/20
$${your}\:{answer}\:{is}\:{correct}. \\ $$$$ \\ $$$${i}\:{asked}\:{an}\:{other}\:{question}:\:{when}\:{the} \\ $$$${dirextrix}\:{is}\:{y}+\mathrm{2}{x}+\mathrm{1}=\mathrm{0},\:{what}\:{is}\:{then} \\ $$$${the}\:{eqn}.\:{of}\:{the}\:{parabola}? \\ $$
Commented by john santu last updated on 10/Apr/20
$${if}\:{focus}\:{at}\:\left(\mathrm{2},\mathrm{5}\right)\:{and}\:{dirextrix}\:{is} \\ $$$${y}+\mathrm{2}{x}+\mathrm{1}\:=\:\mathrm{0},\:{we}\:{get}\:{p}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mid\frac{\mathrm{5}+\mathrm{4}+\mathrm{1}}{\:\sqrt{\mathrm{5}}}\mid \\ $$$${p}\:=\:\sqrt{\mathrm{5}}\:,\:{the}\:{line}\:{passes}\:{through} \\ $$$${focus}\:{orthogonal}\:{to}\:{dirextrix} \\ $$$$\Rightarrow{x}−\mathrm{2}{y}\:+\:\mathrm{6}=\mathrm{0}\: \\ $$$${the}\:{parabola}\:\mathrm{4}{p}\left({x}−{x}_{{v}} \right)=\:\left({y}−{y}_{{v}} \right)^{\mathrm{2}} \\ $$$${by}\:{rotated}\:\theta\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right).\:{it}\:{correct} \\ $$
Commented by som(math1967) last updated on 10/Apr/20
$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} =\frac{\left(\mathrm{2}{x}+{y}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }\: \\ $$
Commented by mr W last updated on 10/Apr/20
$${focus}\:{at}\:\left(\mathrm{2},\mathrm{5}\right) \\ $$$${dirextrix}\:\mathrm{2}{x}+{y}+\mathrm{1}=\mathrm{0} \\ $$$${eqn}.\:{of}\:{parabola}\:{acc}.\:{to}\:{definition}: \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} =\left(\frac{\mathrm{2}{x}+{y}+\mathrm{1}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{xy}−\mathrm{24}{x}−\mathrm{52}{y}+\mathrm{144}=\mathrm{0} \\ $$
Commented by mr W last updated on 10/Apr/20
Commented by Rio Michael last updated on 10/Apr/20
$$\mathrm{wow}\:\mathrm{am}\:\mathrm{so}\:\mathrm{grateful}\:\mathrm{sirs} \\ $$
Commented by john santu last updated on 10/Apr/20
$${how}\:{if}\:{the}\:{vertex}\:{parabola}\:{is}\: \\ $$$$\left(\mathrm{2},−\mathrm{1}\right)\:{and}\:{dirextrix}\:{y}−\mathrm{2}{x}+\mathrm{2}=\mathrm{0}\:? \\ $$$$ \\ $$
Commented by jagoll last updated on 10/Apr/20
$$\mathrm{great}\:\mathrm{sir} \\ $$