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

$$\mathrm{Find}\mathrm{equation}\mathrm{of}\mathrm{an}\mathrm{ellipse}\mathrm{whose}\mathrm{major}\mathrm{axis}$$$$\mathrm{is}\mathrm{vertical},\mathrm{with}\mathrm{the}\mathrm{center}\mathrm{located}(-1,3)$$$$\mathrm{at}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{center}\mathrm{and}\mathrm{one}$$$$\mathrm{of}\mathrm{the}\mathrm{covertices}\mathrm{equal}\mathrm{to}4,\mathrm{and}\mathrm{the}\mathrm{distance}$$$$\mathrm{between}\mathrm{the}\mathrm{center}\mathrm{and}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{vertices}$$$$\mathrm{equal}\mathrm{to}6.$$

Answered by sandy_suhendra last updated on 29/Dec/16

$$\mathrm{distance}\mathrm{between}\mathrm{centre}\mathrm{and}\mathrm{co}\mathrm{vertices}=\mathrm{b}=4$$$$\mathrm{distance}\mathrm{berween}\mathrm{centre}\mathrm{and}\mathrm{vertices}=\mathrm{a}=6$$$$\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{ellipse}=(\mathrm{h},\mathrm{k})=(-1,3)$$$$\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{vertical}\mathrm{ellipse}\equiv \frac{{(\mathrm{x}-\mathrm{h})}^2}{{\mathrm{b}}^2}+\frac{{(\mathrm{y}-\mathrm{k})}^2}{{\mathrm{a}}^2}=1$$$$\frac{{(\mathrm{x}+1)}^2}{4^2}+\frac{{(\mathrm{y}-3)}^2}{6^2}=1\Rightarrow \frac{{(\mathrm{x}+1)}^2}{16}+\frac{{(\mathrm{y}-3)}^2}{36}=1$$

Commented by tawakalitu last updated on 29/Dec/16

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}$$

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