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Question Number 9163 by tawakalitu last updated on 21/Nov/16
show that the ellipse with e = ((√5)/3), focus (0, 2) and directrix x = −((4(√5))/3) has the equation : (((x − (√5))^2 )/9) + (((y − 2)^2 )/4) = 1
$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{ellipse}\:\mathrm{with}\:\mathrm{e}\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{3}},\: \\ $$$$\mathrm{focus}\:\left(\mathrm{0},\:\mathrm{2}\right)\:\mathrm{and}\:\mathrm{directrix}\:\mathrm{x}\:=\:−\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{equation}\::\:\frac{\left(\mathrm{x}\:−\:\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{\left(\mathrm{y}\:−\:\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{1} \\ $$
Commented by sandy_suhendra last updated on 22/Nov/16
equation of an ellipse is (((x−h)^2 )/a^2 ) + (((y−k)^2 )/b^2 ) = 1 where a=major axis b=minor axis (h,k) is the centre of ellipse linear eccentricity=c=(√(a^2 −b^2 )) eccentricity=e=(c/a) directrix ⇒ x=h ± (a/e) focal points : (h±c , k) e = ((√5)/3) ⇒ ((√5)/3) = (c/a) ⇒ c=((a(√5))/3) focus : (0,2) ⇒ h−c=0 ⇒h=c=((a(√5))/3) (h+c is imposible = 0) k=2 directrix ⇒ x=−((4(√5))/5) ( I think that is 5 not 3 ) −((4(√5))/5)=h−(a/e) −((4(√5))/5)=((a(√5))/3)−(a/( (√5)/3)) −((4(√5))/5)=((a(√5))/3)−((3a(√5))/5) (multiplied by ((15)/( (√5)))) −12=5a−9a a=3 ⇒ c=h=(√5) c^2 =a^2 −b^2 5=9−b^2 b=2 substitute : a=3, b=2, h=(√5), k=2 to the equation of ellipse, so we have : (((x−(√5))^2 )/9)+(((y−2)^2 )/4)=1
$$\mathrm{equation}\:\mathrm{of}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{is}\:\:\frac{\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\:\frac{\left(\mathrm{y}−\mathrm{k}\right)^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\mathrm{where}\: \\ $$$$\mathrm{a}=\mathrm{major}\:\mathrm{axis} \\ $$$$\mathrm{b}=\mathrm{minor}\:\mathrm{axis} \\ $$$$\left(\mathrm{h},\mathrm{k}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{ellipse} \\ $$$$\mathrm{linear}\:\mathrm{eccentricity}=\mathrm{c}=\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} } \\ $$$$\mathrm{eccentricity}=\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}} \\ $$$$\mathrm{directrix}\:\Rightarrow\:\mathrm{x}=\mathrm{h}\:\pm\:\frac{\mathrm{a}}{\mathrm{e}} \\ $$$$\mathrm{focal}\:\mathrm{points}\::\:\left(\mathrm{h}\pm\mathrm{c}\:,\:\mathrm{k}\right) \\ $$$$ \\ $$$$\mathrm{e}\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\:\Rightarrow\:\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\:=\:\frac{\mathrm{c}}{\mathrm{a}}\:\Rightarrow\:\mathrm{c}=\frac{\mathrm{a}\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\mathrm{focus}\::\:\left(\mathrm{0},\mathrm{2}\right)\:\Rightarrow\:\mathrm{h}−\mathrm{c}=\mathrm{0}\:\Rightarrow\mathrm{h}=\mathrm{c}=\frac{\mathrm{a}\sqrt{\mathrm{5}}}{\mathrm{3}}\:\left(\mathrm{h}+\mathrm{c}\:\mathrm{is}\:\mathrm{imposible}\:=\:\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{k}=\mathrm{2} \\ $$$$ \\ $$$$\mathrm{directrix}\:\Rightarrow\:\mathrm{x}=−\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{5}}\:\left(\:\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{is}\:\mathrm{5}\:\mathrm{not}\:\mathrm{3}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{5}}=\mathrm{h}−\frac{\mathrm{a}}{\mathrm{e}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{5}}=\frac{\mathrm{a}\sqrt{\mathrm{5}}}{\mathrm{3}}−\frac{\mathrm{a}}{\:\sqrt{\mathrm{5}}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{5}}=\frac{\mathrm{a}\sqrt{\mathrm{5}}}{\mathrm{3}}−\frac{\mathrm{3a}\sqrt{\mathrm{5}}}{\mathrm{5}}\:\:\left(\mathrm{multiplied}\:\mathrm{by}\:\frac{\mathrm{15}}{\:\sqrt{\mathrm{5}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{12}=\mathrm{5a}−\mathrm{9a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}=\mathrm{3}\:\Rightarrow\:\mathrm{c}=\mathrm{h}=\sqrt{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{5}=\mathrm{9}−\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{substitute}\::\:\mathrm{a}=\mathrm{3},\:\mathrm{b}=\mathrm{2},\:\mathrm{h}=\sqrt{\mathrm{5}},\:\mathrm{k}=\mathrm{2} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{ellipse},\:\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:: \\ $$$$\frac{\left(\mathrm{x}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{9}}+\frac{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1} \\ $$
Commented by tawakalitu last updated on 22/Nov/16
Wow, this is great. i really appreciate sir. God bless you.
$$\mathrm{Wow},\:\mathrm{this}\:\mathrm{is}\:\mathrm{great}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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