Question

Put the equation 169x^2+25y^2-4225=0 in standard form and sketch the ellipse including the foci in your sketch ​

Answer 1

The standard form of the ellipse is
`(x^(2))/(25) +(y^(2))/(169) = 1`. Coordinates of the foci are
`F_(1) (x,y) = (0,-c)` and
`F_(2)(x,y) = (0,c)`, respectively.

Explanation:

The equation of a ellipse centered in the origin in standard form is defined by following formula:

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

Where
`a, b` are the coefficients of the ellipse.

Now we proceed to transform the equation of the ellipse in general form into standard form by algebraic means:

`169\cdot x^(2)+25\cdot y^(2)-4225 = 0`

`169\cdot x^(2)+25\cdot y^(2) = 4225`

`(x^(2))/(25) +(y^(2))/(169) = 1`

Since
`b > a`, then major axis of the ellipse is located in y-axis. The distance between center and focus (
`c`) is calculated by following Pythagorean identity:

`c =\sqrt{b^(2)-a^(2)}` (2)

`c = 12`

The location of the foci are represented by
`F_(1) (x,y) = (0,-c)` and
`F_(2)(x,y) = (0,c)`. If we know that
`c = 12`, then the location of the foci are, respectively:

`F_(1)(x,y) = (0,-12)`,
`F_(2)(x,y) = (0,12)`