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प्रश्न
Identify the type of conic and find centre, foci, vertices, and directrices of the following:
18x2 + 12y2 – 144x + 48y + 120 = 0
उत्तर
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`
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