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Question Number 136165 by bramlexs22 last updated on 19/Mar/21

Find the equation of ellipse with F_1 (1,2) , F_2 (3,4) and 2a = 2(√3)

$${Find}\:{the}\:{equation}\:{of}\:{ellipse}\:{with} \\ $$$${F}_{\mathrm{1}} \left(\mathrm{1},\mathrm{2}\right)\:,\:{F}_{\mathrm{2}} \left(\mathrm{3},\mathrm{4}\right)\:{and}\:\mathrm{2}{a}\:=\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$

Commented by bramlexs22 last updated on 19/Mar/21

Dear Mr W can you help me

$${Dear}\:{Mr}\:{W}\:{can}\:{you}\:{help}\:{me} \\ $$

Commented by mr W last updated on 19/Mar/21

do you mean the foci with F_1 and F_2 ?

$${do}\:{you}\:{mean}\:{the}\:{foci}\:{with}\:{F}_{\mathrm{1}} \:{and}\:{F}_{\mathrm{2}} ? \\ $$

Commented by bramlexs22 last updated on 19/Mar/21

yes sir

$${yes}\:{sir} \\ $$

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 2a. (√((x−1)^2 +(y−2)^2 ))+(√((x−3)^2 +(y−4)^2 ))=2(√3) (√((x−1)^2 +(y−2)^2 ))=2(√3)−(√((x−3)^2 +(y−4)^2 )) (x−1)^2 +(y−2)^2 =12+(x−3)^2 +(y−4)^2 −4(√3)(√((x−3)^2 +(y−4)^2 )) x+y−8=(√3)(√((x−3)^2 +(y−4)^2 )) x^2 +y^2 +64+2xy−16x−16y=3(x^2 −6x+9)+3(y^2 −8y+16) ⇒2x^2 +2y^2 −2xy−2x−8y+11=0

$${per}\:{definition}\:{the}\:{sum}\:{of}\:{distances} \\ $$$${from}\:{any}\:{point}\:{on}\:{an}\:{ellipse}\:{to}\:{the} \\ $$$${foci}\:{is}\:{constant}\:{which}\:{is}\:\mathrm{2}{a}. \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{3}}−\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{12}+\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$${x}+{y}−\mathrm{8}=\sqrt{\mathrm{3}}\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{64}+\mathrm{2}{xy}−\mathrm{16}{x}−\mathrm{16}{y}=\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)+\mathrm{3}\left({y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{16}\right) \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{2}{x}−\mathrm{8}{y}+\mathrm{11}=\mathrm{0} \\ $$

Commented by bramlexs22 last updated on 19/Mar/21

i try by rotation with angle 45° . i′m stuck. forgot by definition. thanks sir

$${i}\:{try}\:{by}\:{rotation}\:{with}\:{angle} \\ $$$$\mathrm{45}°\:.\:{i}'{m}\:{stuck}.\:{forgot}\:{by}\:{definition}.\:{thanks}\:{sir} \\ $$

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 45°.

$${in}\:{this}\:{case}\:{you}\:{can}\:{also}\:{easily}\:{get} \\ $$$${the}\:{equation}\:{through}\:{rotation}, \\ $$$${because}\:{the}\:{rotation}\:{angle}\:{is}\:\mathrm{45}°. \\ $$

Commented by bramlexs22 last updated on 19/Mar/21

yes sir but i′m stuck. i don′t which my mistake

$${yes}\:{sir}\:{but}\:{i}'{m}\:{stuck}.\:{i}\:{don}'{t} \\ $$$${which}\:{my}\:{mistake} \\ $$

Commented by bramlexs22 last updated on 19/Mar/21

can sir show me my mistake

can sir show me my mistake

Commented by bramlexs22 last updated on 19/Mar/21

((x),(y) ) = (((cos 45° −sin 45°)),((sin 45° cos 45°)) )^(−1) (((x′)),((y′)) )

$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{cos}\:\mathrm{45}°\:\:\:−\mathrm{sin}\:\mathrm{45}°}\\{\mathrm{sin}\:\mathrm{45}°\:\:\:\:\:\:\:\:\mathrm{cos}\:\mathrm{45}°}\end{pmatrix}^{−\mathrm{1}} \:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix} \\ $$

Commented by mr W last updated on 19/Mar/21

Method using rotation F_1 F_2 =(√((3−1)^2 +(4−2)^2 ))=2(√2) midpoint from F_1 and F_2 is (2,3) a=(√3) (given) b=(√(a^2 −(((F_1 F_2 )/2))^2 ))=(√(3−2))=1 eqn. before rotation: (((x−2)^2 )/3)+(y−3)^2 =1 rotation by 45° ↶ about point (2,3) x_1 =(x−2)((√2)/2)+(y−3)((√2)/2)+2=(x+y−5)((√2)/2)+2 y_1 =−(x−2)((√2)/2)+(y−3)((√2)/2)+3=(−x+y−1)((√2)/2)+3 eqn. after rotation: (((x+y−5)^2 )/6)+(((−x+y−1)^2 )/2)=1 or (x+y−5)^2 +3(−x+y−1)^2 =6 x^2 +y^2 +25+2xy−10x−10y+3(x^2 +y^2 +1−2xy+2x−2y)=6 ⇒2x^2 +2y^2 −2xy−2x−8y+11=0

$$\boldsymbol{{Method}}\:\boldsymbol{{using}}\:\boldsymbol{{rotation}} \\ $$$${F}_{\mathrm{1}} {F}_{\mathrm{2}} =\sqrt{\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${midpoint}\:{from}\:{F}_{\mathrm{1}} \:{and}\:{F}_{\mathrm{2}} \:{is}\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${a}=\sqrt{\mathrm{3}}\:\left({given}\right) \\ $$$${b}=\sqrt{{a}^{\mathrm{2}} −\left(\frac{{F}_{\mathrm{1}} {F}_{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{3}−\mathrm{2}}=\mathrm{1} \\ $$$${eqn}.\:{before}\:{rotation}: \\ $$$$\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}}+\left({y}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${rotation}\:{by}\:\mathrm{45}°\:\curvearrowleft\:{about}\:{point}\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${x}_{\mathrm{1}} =\left({x}−\mathrm{2}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\left({y}−\mathrm{3}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{2}=\left({x}+{y}−\mathrm{5}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{2} \\ $$$${y}_{\mathrm{1}} =−\left({x}−\mathrm{2}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\left({y}−\mathrm{3}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{3}=\left(−{x}+{y}−\mathrm{1}\right)\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{3} \\ $$$${eqn}.\:{after}\:{rotation}: \\ $$$$\frac{\left({x}+{y}−\mathrm{5}\right)^{\mathrm{2}} }{\mathrm{6}}+\frac{\left(−{x}+{y}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{1} \\ $$$${or} \\ $$$$\left({x}+{y}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{3}\left(−{x}+{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{6} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{25}+\mathrm{2}{xy}−\mathrm{10}{x}−\mathrm{10}{y}+\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{xy}+\mathrm{2}{x}−\mathrm{2}{y}\right)=\mathrm{6} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{xy}−\mathrm{2}{x}−\mathrm{8}{y}+\mathrm{11}=\mathrm{0} \\ $$

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 (0,0) sir. great sir

$${oo}\:{my}\:{mistake}\:{use}\:{rotatiton}\: \\ $$$${about}\:{point}\:\left(\mathrm{0},\mathrm{0}\right)\:{sir}.\:{great}\:{sir} \\ $$

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