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Question: The centre of \(14x^{2} - 4xy + 11y^{2} - 44x - 58y + 71 = 0\)is...

The centre of 14x2−4xy+11y2−44x−58y+71=014x^{2} - 4xy + 11y^{2} - 44x - 58y + 71 = 0is

A

(2, 3)

B

(2, –3)

C

(–2, 3)

D

(–2, –3)

Answer

(2, 3)

Explanation

Solution

Centre of conic is (hf−bgab−h2,gh−afab−h2)\left( \frac{hf - bg}{ab - h^{2}},\frac{gh - af}{ab - h^{2}} \right)

Here, a=14,h=−2a = 14,h = - 2, b=11b = 11, g=−22g = - 22, f=−29f = - 29, c=71c = 71Centre ≡((−2)(−29)−(11)(−22)(14)(11)−(−2)2,(−22)(−2)−(14)(−29)(14)(11)−(−2)2)\equiv \left( \frac{( - 2)( - 29) - (11)( - 22)}{(14)(11) - ( - 2)^{2}},\frac{( - 22)( - 2) - (14)( - 29)}{(14)(11) - ( - 2)^{2}} \right)Centre ≡(2,3)\equiv (2,3)