Question Number 47480 by ajfour last updated on 10/Nov/18
Answered by MrW3 last updated on 10/Nov/18
![assume the major and minor axes are parallel to x axis and y axis. eqn. (((x−h)^2 )/u^2 )+(((y−k)^2 )/v^2 )=1 h=((a+c)/2) k=((b+d)/2) (((a−h)^2 )/u^2 )+(((−k)^2 )/v^2 )=1 ...(i) (((−h)^2 )/u^2 )+(((b−k)^2 )/v^2 )=1 ...(ii) (((a−h)^2 (b−k)^2 −h^2 k^2 )/u^2 )=(b−k)^2 −k^2 (1/u^2 )=(((b−k)^2 −k^2 )/((a−h)^2 (b−k)^2 −h^2 k^2 ))=((b(b−2k))/([(a−h)(b−k)+hk][(a−h)(b−k)−hk])) ⇒(1/u^2 )=((4bd)/((ab+cd)(bc+ad))) (((b−k)^2 (a−h)^2 −h^2 k^2 )/v^2 )=(a−h)^2 −h^2 (1/v^2 )=(((a−h)^2 −h^2 )/((b−k)^2 (a−h)^2 −h^2 k^2 ))=((a(a−2h))/([(a−h)(b−k)+hk][(a−h)(b−k)−hk])) ⇒(1/v^2 )=((4ac)/((ab+cd)(bc+ad))) eqn. of ellipse: (((2x−a−c)^2 bd)/((ab+cd)(bc+ad)))+(((2y−b−d)^2 ac)/((ab+cd)(bc+ad)))=1 or ⇒(2x−a−c)^2 bd+(2y−b−d)^2 ac=(ab+cd)(bc+ad)](https://www.tinkutara.com/question/Q47488.png)
$${assume}\:{the}\:{major}\:{and}\:{minor}\:{axes} \\ $$$${are}\:{parallel}\:{to}\:{x}\:{axis}\:{and}\:{y}\:{axis}. \\ $$$${eqn}. \\ $$$$\frac{\left({x}−{h}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} }+\frac{\left({y}−{k}\right)^{\mathrm{2}} }{{v}^{\mathrm{2}} }=\mathrm{1} \\ $$$${h}=\frac{{a}+{c}}{\mathrm{2}} \\ $$$${k}=\frac{{b}+{d}}{\mathrm{2}} \\ $$$$\frac{\left({a}−{h}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} }+\frac{\left(−{k}\right)^{\mathrm{2}} }{{v}^{\mathrm{2}} }=\mathrm{1}\:\:\:…\left({i}\right) \\ $$$$\frac{\left(−{h}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} }+\frac{\left({b}−{k}\right)^{\mathrm{2}} }{{v}^{\mathrm{2}} }=\mathrm{1}\:\:\:…\left({ii}\right) \\ $$$$\frac{\left({a}−{h}\right)^{\mathrm{2}} \left({b}−{k}\right)^{\mathrm{2}} −{h}^{\mathrm{2}} {k}^{\mathrm{2}} }{{u}^{\mathrm{2}} }=\left({b}−{k}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{u}^{\mathrm{2}} }=\frac{\left({b}−{k}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} }{\left({a}−{h}\right)^{\mathrm{2}} \left({b}−{k}\right)^{\mathrm{2}} −{h}^{\mathrm{2}} {k}^{\mathrm{2}} }=\frac{{b}\left({b}−\mathrm{2}{k}\right)}{\left[\left({a}−{h}\right)\left({b}−{k}\right)+{hk}\right]\left[\left({a}−{h}\right)\left({b}−{k}\right)−{hk}\right]} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{u}^{\mathrm{2}} }=\frac{\mathrm{4}{bd}}{\left({ab}+{cd}\right)\left({bc}+{ad}\right)} \\ $$$$ \\ $$$$\frac{\left({b}−{k}\right)^{\mathrm{2}} \left({a}−{h}\right)^{\mathrm{2}} −{h}^{\mathrm{2}} {k}^{\mathrm{2}} }{{v}^{\mathrm{2}} }=\left({a}−{h}\right)^{\mathrm{2}} \:−{h}^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{v}^{\mathrm{2}} }=\frac{\left({a}−{h}\right)^{\mathrm{2}} \:−{h}^{\mathrm{2}} }{\left({b}−{k}\right)^{\mathrm{2}} \left({a}−{h}\right)^{\mathrm{2}} −{h}^{\mathrm{2}} {k}^{\mathrm{2}} }=\frac{{a}\left({a}−\mathrm{2}{h}\right)}{\left[\left({a}−{h}\right)\left({b}−{k}\right)+{hk}\right]\left[\left({a}−{h}\right)\left({b}−{k}\right)−{hk}\right]} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{v}^{\mathrm{2}} }=\frac{\mathrm{4}{ac}}{\left({ab}+{cd}\right)\left({bc}+{ad}\right)} \\ $$$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{\left(\mathrm{2}{x}−{a}−{c}\right)^{\mathrm{2}} {bd}}{\left({ab}+{cd}\right)\left({bc}+{ad}\right)}+\frac{\left(\mathrm{2}{y}−{b}−{d}\right)^{\mathrm{2}} {ac}}{\left({ab}+{cd}\right)\left({bc}+{ad}\right)}=\mathrm{1} \\ $$$${or} \\ $$$$\Rightarrow\left(\mathrm{2}{x}−{a}−{c}\right)^{\mathrm{2}} {bd}+\left(\mathrm{2}{y}−{b}−{d}\right)^{\mathrm{2}} {ac}=\left({ab}+{cd}\right)\left({bc}+{ad}\right) \\ $$
Commented by ajfour last updated on 10/Nov/18
$${Thank}\:{you}\:{Sir}.\:\left({Matches}!\right) \\ $$