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kudzordzifrancis

Answer:

The foci are located at: [tex](-\sqrt{13},0)[/tex]  and  [tex](\sqrt{13},0)[/tex]

Step-by-step explanation:

The given conic has equation: [tex]36x^2+49y^2=1764[/tex]

Divide each term by 1764

[tex]\frac{36x^2}{1764}+\frac{49y^2}{1764}=\frac{1764}{1764}[/tex]

[tex]\implies \frac{x^2}{49}+\frac{y^2}{36}=1[/tex]

[tex]\implies \frac{x^2}{7^2}+\frac{y^2}{6^2}=1[/tex]

Comparing to:

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

We have a=7 and b=6

We use the relation [tex]a^2-b^2=c^2[/tex] to find c.

[tex]\implies 7^2-6^2=c^2[/tex]

[tex]\implies 49-36=c^2[/tex]

[tex]\implies 13=c^2[/tex]

[tex]\implies c=\pm \sqrt{13}[/tex]

The foci are located at: [tex](-\sqrt{13},0)[/tex]  and  [tex](\sqrt{13},0)[/tex]

absor201

Answer:

Foci of given equation located at  (-√13, 0) and (√13,0) ....

Step-by-step explanation:

We have to find the foci point:

The equation is:

36x2 + 49y2 = 1,764

We can write the equation as:

x^2/49 + y^2/36 = 1

OR

Divide the equation by 1764

36x2/1764+ 49y2/1764 = 1,764/1764

x^2/49 + y^2/36 = 1

It is a horizontal ellipse.

(x-h)^2/ a^2 + (y-k)^2/b^2 = 1

The foci of ellipse are (h+c,k) and (h-c,k) where c^2= a^2-b^2

The above equation is:

x^2/49 + y^2/36 = 1

1/(7)^2 x^2 + 1/(6)^2 y^2 = 1

By comparing the above equation we get;

a=7 , b = 6 , h = k = 0

c^2 = a^2 - b^2

c^2 = (7)^2 - (6)^2

c^2 = 49 - 36

c^2 = 13

Take square root at both sides:

√c^2 = +/-√13

c = +/-√13

Thus,  Foci of given equation located at  (-√13, 0) and (√13,0) ....

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