Question Number 136165 by bramlexs22 last updated on 19/Mar/21
$${Find}\:{the}\:{equation}\:{of}\:{ellipse}\:{with} \\ $$$${F}_{\mathrm{1}} \left(\mathrm{1},\mathrm{2}\right)\:,\:{F}_{\mathrm{2}} \left(\mathrm{3},\mathrm{4}\right)\:{and}\:\mathrm{2}{a}\:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$
Commented by bramlexs22 last updated on 19/Mar/21
$${Dear}\:{Mr}\:{W}\:{can}\:{you}\:{help}\:{me} \\ $$
Commented by mr W last updated on 19/Mar/21
$${do}\:{you}\:{mean}\:{the}\:{foci}\:{with}\:{F}_{\mathrm{1}} \:{and}\:{F}_{\mathrm{2}} ? \\ $$
Commented by bramlexs22 last updated on 19/Mar/21
$${yes}\:{sir} \\ $$
Answered by mr W last updated on 19/Mar/21
$${per}\:{definition}\:{the}\:{sum}\:{of}\:{distances} \\ $$$${from}\:{any}\:{point}\:{on}\:{an}\:{ellipse}\:{to}\:{the} \\ $$$${foci}\:{is}\:{constant}\:{which}\:{is}\:\mathrm{2}{a}. \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{3}}−\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{12}+\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$${x}+{y}−\mathrm{8}=\sqrt{\mathrm{3}}\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{64}+\mathrm{2}{xy}−\mathrm{16}{x}−\mathrm{16}{y}=\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)+\mathrm{3}\left({y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{16}\right) \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{2}{x}−\mathrm{8}{y}+\mathrm{11}=\mathrm{0} \\ $$
Commented by bramlexs22 last updated on 19/Mar/21
$${i}\:{try}\:{by}\:{rotation}\:{with}\:{angle} \\ $$$$\mathrm{45}°\:.\:{i}'{m}\:{stuck}.\:{forgot}\:{by}\:{definition}.\:{thanks}\:{sir} \\ $$
Commented by mr W last updated on 19/Mar/21
Commented by mr W last updated on 19/Mar/21
$${in}\:{this}\:{case}\:{you}\:{can}\:{also}\:{easily}\:{get} \\ $$$${the}\:{equation}\:{through}\:{rotation}, \\ $$$${because}\:{the}\:{rotation}\:{angle}\:{is}\:\mathrm{45}°. \\ $$
Commented by bramlexs22 last updated on 19/Mar/21
$${yes}\:{sir}\:{but}\:{i}'{m}\:{stuck}.\:{i}\:{don}'{t} \\ $$$${which}\:{my}\:{mistake} \\ $$
Commented by bramlexs22 last updated on 19/Mar/21
can sir show me my mistake
Commented by bramlexs22 last updated on 19/Mar/21
$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{cos}\:\mathrm{45}°\:\:\:−\mathrm{sin}\:\mathrm{45}°}\\{\mathrm{sin}\:\mathrm{45}°\:\:\:\:\:\:\:\:\mathrm{cos}\:\mathrm{45}°}\end{pmatrix}^{−\mathrm{1}} \:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix} \\ $$
Commented by mr W last updated on 19/Mar/21
$$\boldsymbol{{Method}}\:\boldsymbol{{using}}\:\boldsymbol{{rotation}} \\ $$$${F}_{\mathrm{1}} {F}_{\mathrm{2}} =\sqrt{\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${midpoint}\:{from}\:{F}_{\mathrm{1}} \:{and}\:{F}_{\mathrm{2}} \:{is}\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${a}=\sqrt{\mathrm{3}}\:\left({given}\right) \\ $$$${b}=\sqrt{{a}^{\mathrm{2}} −\left(\frac{{F}_{\mathrm{1}} {F}_{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{3}−\mathrm{2}}=\mathrm{1} \\ $$$${eqn}.\:{before}\:{rotation}: \\ $$$$\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}}+\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${rotation}\:{by}\:\mathrm{45}°\:\curvearrowleft\:{about}\:{point}\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${x}_{\mathrm{1}} =\left({x}−\mathrm{2}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\left({y}−\mathrm{3}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{2}=\left({x}+{y}−\mathrm{5}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{2} \\ $$$${y}_{\mathrm{1}} =−\left({x}−\mathrm{2}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\left({y}−\mathrm{3}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{3}=\left(−{x}+{y}−\mathrm{1}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{3} \\ $$$${eqn}.\:{after}\:{rotation}: \\ $$$$\frac{\left({x}+{y}−\mathrm{5}\right)^{\mathrm{2}} }{\mathrm{6}}+\frac{\left(−{x}+{y}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{1} \\ $$$${or} \\ $$$$\left({x}+{y}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{3}\left(−{x}+{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{6} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{25}+\mathrm{2}{xy}−\mathrm{10}{x}−\mathrm{10}{y}+\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{xy}+\mathrm{2}{x}−\mathrm{2}{y}\right)=\mathrm{6} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{2}{x}−\mathrm{8}{y}+\mathrm{11}=\mathrm{0} \\ $$
Commented by mr W last updated on 19/Mar/21
Commented by bramlexs22 last updated on 19/Mar/21
$${oo}\:{my}\:{mistake}\:{use}\:{rotatiton}\: \\ $$$${about}\:{point}\:\left(\mathrm{0},\mathrm{0}\right)\:{sir}.\:{great}\:{sir} \\ $$