Question

Write the equation of the conic `18y^2-5x^2+72y+30x-63=0` in standard form?

Answer 1

`(y+2)^2/(sqrt2)^2-(x-3)^2/(6/sqrt5)^2=1`

Explanation:

This is the equation of typical hyperbola whose conjugate axis is parallel to `y`-axis. Its standard form is

`(y-k)^2/a^2-(x-h)^2/b^2=1` - whose center is `(h,k)`, major axis is `y=k` and conjugate axis is `x=h`.

We convert it into standard form as follows:

`18y^2-5x^2+72y+30x-63=0`

or `18(y^2+4y)-5(x^2-6x)=63`

or `18(y^2+4y+4)-5(x^2-6x+9)-72+45=63`

or `18(y+2)^2-5(x-3)^2=36`

or `(y+2)^2/2-(x-3)^2/(36/5)=1`

or `(y+2)^2/(sqrt2)^2-(x-3)^2/(6/sqrt5)^2=1`

and hence center is `(3,-2)`, major axis is `y=-2` and conjugate axis is `x=3`. ts graph appears as shown below:

graph{(18y^2-5x^2+72y+30x-63)((x-3)^2+(y+2)^2-0.03)(y+2)(x-3)=0 [-18.33, 21.67, -12.32, 7.68]}