Question 1043225
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Find the center,foci,vertices and covertices of the ellipse with the given equation.

9x^2+16y^2-126x+64y=71
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<pre>
{{{9x^2+16y^2-126x+64y}}} = {{{71}}}  --->

{{{9*(x^2-14x) + 16*(y^2+4y)}}} = {{{71}}}  --->

{{{9*(x-7)^2 - 9*49 + 16*(y+2)^2 - 16*4}}} = {{{171}}},

{{{9*(x-7)^2 + 16*(y+2)^2}}} = {{{171+9*49 + 16*4}}},

{{{9*(x-7)^2 + 16*(y+2)^2}}} = {{{171+9*49 + 16*4}}},

{{{9*(x-7)^2 + 16*(y+2)^2}}} = {{{676}}},      ( <--- next divide both sides by 676 = {{{26^2}}} )

{{{(x-7)^2/(26/3)^2 + (y+2)^2/(26/4)^2}}} = 1,  or

{{{(x-7)^2/(26/3)^2 + (y+2)^2/(13/2)^2}}} = 1.

The center of the ellipse is the point  (7,-2).

The semi-major axis is a = {{{26/3}}}.

The semi-minor axis is b = {{{26/4}}} = 13/2.   (Notice that a > b. The major axis is horizontal, and the ellipse is wider than tall).

Two vertices in horizontal directions are  ({{{7+(26/3)}}},{{{-2}}})  and  ({{{7-(26/3)}}},{{{-2}}}).

Two vertices in vertical directions are  ({{{7}}},{{{-2+(13/2)}}})  and  ({{{7}}},{{{-2-(13/2)}}}).

The linear eccentricity is c = {{{sqrt(a^2-b^2)}}} = {{{sqrt((26/3)^2-(26/4)^2)}}} = {{{26*sqrt((1/9)-(1/16))}}} = {{{26*sqrt((16-9)/(9*16))}}} = {{{(26/(3*4))*sqrt(7)}}} = {{{(13/6)*sqrt(7)}}}.

The foci are  ({{{7+(13/6)sqrt(7)}}},{{{-2}}})  and  ({{{7-(13/6)sqrt(7)}}},{{{-2}}}).
</pre>

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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Ellipse-definition--canonical-equation--characteristic-points-and-elements.lesson>Ellipse definition, canonical equation, characteristic points and elements</A>

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