Question

Identify the type of conic and find centre, foci, vertices, and directrices of the following:

18x² + 12y² – 144x + 48y + 120 = 0

Answer 1

18x² – 144x + 12y² + 48y = – 120

18(x² – 8x) + 12(y² + 4y) = – 120

18(x – 4)² + 12(y + 2)² = – 120 + 288 + 48

18(x – 4)² + 12(y + 2)² = 216

`(18(x - 4)^2)/216 + (12(y + 2)^2)/216` = 1

`(x - 4)^2/12 + (y + 2)^2/18` = 1

It is an ellipse.

The major axis is parallell to y axis.

a² = 18, b² = 12

a = `sqrt(18), "b" = 2sqrt(3)`

= `3sqrt(2)`

c² = a² + b²

= 18 – 12

= 6

c = `sqrt(6)`

ae = `sqrt(6)`

`3sqrt(2)"e" = sqrt(6)`

e = `sqrt(6)/(3sqrt(2)) = sqrt(3)/3 = 1/sqrt(3)`

Centre (h, k) = (4, – 2)

Vertices (h, ±a + k) = `(4, +-  3sqrt(2) - 2)`

= `(4, 3sqrt(2) - 2)` and `(4, -3sqrt(2) - 2)`

Foci (h, ±c + k) = `(4, +- sqrt(6) - 2)`

= `(4 +-  sqrt(6) - 2)` and `(4, - sqrt(6) - 2)`

Directrix y = `+-  "a"/"e" + "k"`

= `+-  (3sqrt(2))/1 - 2`

= `+-  3sqrt(6) - 2`

y = `3sqrt(6) - 2` and y = `- 3sqrt(6) - 2`