Question

Question #30429

Answer 1

The standard form is `(x+1)^2/(sqrt5)^2+(y-1)^2/4^2=1`

Explanation:

Let's rearrange the equation by completing the squares

`16x^2+5y^2+32x-10y-59=0`

`16(x^2+2x)+5(y^2-2y)=59`

`16(x^2+2x+1)+5(y^2-2y+1)=59+16+5`

`16(x+1)^2+5(y-1)^2=80`

`16(x+1)^2/80+5(y-1)^2/80=1`

`(x+1)^2/5+(y-1)^2/16=1`

`(x+1)^2/(sqrt5)^2+(y-1)^2/4^2=1`

This is an ellipse, center `=(-1,1)`

graph{(16x^2+32x+5y^2-10y-59)((x+1)^2+(y-1)^2-0.01)=0 [-11.25, 11.25, -5.625, 5.625]}

Answer 2

Yes

Explanation:

To convert the equation to standard form for ellipses, we need to complete the square.
Note that the standard form for an ellipse is:
`((x-h)^2)/a^2+((y-k)^2)/b^2=1` with a horizontal semi-major axis
or
`((x-h)^2)/b^2+((y-k)^2)/a^2=1` with a vertical semi-major axis
where `(h,k)` is the center of the ellipse and `a>b`, where `a` is the length of the semi-major axis and `b` is the length of the semi-minor axis

Given the starting equation `16x^2+5y^2+32x-10y-59=0`, we first add 59 to both sides to isolate the variables and the constants:
`16x^2+5y^2+32x-10y=59`

Now we can complete the square for x and y:
`16(x^2+2x)+5(y^2-2y)=59`
`16(x+1)^2-16+5(y-1)^2-5=59`
`16(x+1)^2+5(y-1)^2=80`

Now we divide both sides by 80 to make the right-hand side equal to 1 for standard conic form:
`((x+1)^2)/5+((y-1)^2)/16=1`

Therefore, this conic is a vertical ellipse centered at `(-1,1)` with a semi-major axis of length `4` and a semi-minor axis of length `sqrt(5)`.