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Question 80 of 126

Q.The eccentricity of the conic 9x2+5y2−54x−40y+116=09x^2 + 5y^2 - 54x - 40y + 116 = 0 is :

(a) 13\dfrac{1}{3}
(b) 23\dfrac{2}{3}
(c) 49\dfrac{4}{9}
(d) 25\dfrac{2}{\sqrt5}
Puducherry TnboardTamil Nadu HSC (DGE) Board 2017MCQ· 1mImportance★★★★★
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Complete the square in xx and yy to bring the conic to standard ellipse form, identify the larger denominator as a2a^2, and apply e=1−b2/a2e=\sqrt{1-b^2/a^2}.

  1. Given: 9x2+5y2−54x−40y+116=09x^2+5y^2-54x-40y+116=0.
  2. Group and complete the square: 9(x2−6x)+5(y2−8y)+116=09(x^2-6x) + 5(y^2-8y) + 116 = 0.
  3. 9(x2−6x+9−9)+5(y2−8y+16−16)+116=0⇒9(x−3)2−81+5(y−4)2−80+116=09(x^2-6x+9-9) + 5(y^2-8y+16-16) + 116=0 \Rightarrow 9(x-3)^2-81 + 5(y-4)^2-80+116=0.
  4. Simplify constants: −81−80+116=−45-81-80+116=-45, so 9(x−3)2+5(y−4)2−45=0⇒9(x−3)2+5(y−4)2=459(x-3)^2+5(y-4)^2-45=0 \Rightarrow 9(x-3)^2+5(y-4)^2=45.
  5. Divide by 45: (x−3)25+(y−4)29=1\dfrac{(x-3)^2}{5}+\dfrac{(y-4)^2}{9}=1. …

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