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Find-equation-of-an-ellipse-whose-major-axis-is-vertical-with-the-center-located-1-3-at-the-distance-between-the-center-and-one-of-the-covertices-equal-to-4-and-the-distance-between-the-center

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Question Number 9732 by tawakalitu last updated on 29/Dec/16
Find equation of an ellipse whose major axis is vertical, with the center located (− 1, 3) at the distance between the center and one of the covertices equal to 4, and the distance between the center and one of the vertices equal to 6.
$$\mathrm{Find}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{whose}\:\mathrm{major}\:\mathrm{axis} \\ $$$$\mathrm{is}\:\mathrm{vertical},\:\mathrm{with}\:\mathrm{the}\:\mathrm{center}\:\mathrm{located}\:\left(−\:\mathrm{1},\:\mathrm{3}\right) \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{center}\:\mathrm{and}\:\mathrm{one}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{covertices}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{4},\:\mathrm{and}\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{center}\:\mathrm{and}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vertices}\: \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{6}. \\ $$
Answered by sandy_suhendra last updated on 29/Dec/16
distance between centre and co vertices=b=4 distance berween centre and vertices=a=6 the centre of ellipse=(h,k)=(−1,3) the equation of vertical ellipse ≡ (((x−h)^2 )/b^2 )+(((y−k)^2 )/a^2 )=1 (((x+1)^2 )/4^2 )+(((y−3)^2 )/6^2 )=1 ⇒(((x+1)^2 )/(16))+(((y−3)^2 )/(36))=1
$$\mathrm{distance}\:\mathrm{between}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{co}\:\mathrm{vertices}=\mathrm{b}=\mathrm{4} \\ $$$$\mathrm{distance}\:\mathrm{berween}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{vertices}=\mathrm{a}=\mathrm{6} \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{ellipse}=\left(\mathrm{h},\mathrm{k}\right)=\left(−\mathrm{1},\mathrm{3}\right) \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{vertical}\:\mathrm{ellipse}\:\equiv\:\frac{\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }+\frac{\left(\mathrm{y}−\mathrm{k}\right)^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }+\frac{\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{16}}+\frac{\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{36}}=\mathrm{1} \\ $$
Commented by tawakalitu last updated on 29/Dec/16
Thank you sir. God bless you
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

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