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ludmilkaskok [199]
3 years ago
12

I need help like RN!!! what are the foci of the ellipse? graph the ellipse. 18x^2+36y^=648

Mathematics
2 answers:
Thepotemich [5.8K]3 years ago
7 0
Hi,

18x^2+36y^2=648\\

\Rightarrow ( \dfrac{x}{6})^2+( \dfrac{y}{3* \sqrt{2} } )^2=1\\

a=6,\ b=3* \sqrt{2}\\

c= \sqrt{a^2-b^2} = \sqrt{36-18} =3* \sqrt{2}\\


Center is (0,0)
Foci are (-3√2,0) and (3√2,0)

Lerok [7]3 years ago
6 0

Answer:

The equation of ellipse is given by:

\frac{x^2}{a^2}+\frac{y^2}{b^2} =1                 ....[1]

where,

a and b are the semi major axis and semi minor axis.

Foci of ellipse = (\pm c, 0)

where, c = \sqrt{a^2-b^2}

As per the statement:

Given the ellipse is:

18x^2+36y^2 = 648

Divide both sides by 648 we have;

\frac{x^2}{36}+\frac{y^2}{18} =1

On comparing with [1] we have;

a^2 =36 and b^2=18

First find c:

c = \sqrt{36-18}=\sqrt{18}=3\sqrt{2}

Foci of the ellipse are:

(\pm 3\sqrt{2}, 0)

Therefore, the foci of the ellipse are, (3\sqrt{2}, 0) and (-3\sqrt{2} , 0)

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