Question

36x2 + 49y2 = 1,764 The foci are located at

Answer 1

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

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) ....

Answer 2

The foci are located at:
`(-√(13),0)` and
`(√(13),0)`

Explanation:

The given conic has equation:
`36x^2+49y^2=1764`

Divide each term by 1764

`(36x^2)/(1764)+(49y^2)/(1764)=(1764)/(1764)`

`\implies (x^2)/(49)+(y^2)/(36)=1`

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

Comparing to:

`(x^2)/(a^2)+(y^2)/(b^2)=1`

We have a=7 and b=6

We use the relation
`a^2-b^2=c^2` to find c.

`\implies 7^2-6^2=c^2`

`\implies 49-36=c^2`

`\implies 13=c^2`

`\implies c=\pm √(13)`

The foci are located at:
`(-√(13),0)` and
`(√(13),0)`