A curve is represented by c - 3x-3 y-22-23 x-y-12-180- Eccentricity of the curve is A- 31B- 1-3C-D- 2-5

The correct option is B 1/√(3)3(x 3y+3)^2+2(3x+y 1)^2=180 ⇒ (x 3y+3)^2/60 + (3x+y 1)/90=1 ⇒ ( x 3y+3/√(1+3^2)√(6) )^2+ (3x+y 1/√(1+3^2)× 3 )^2 =1 ∴ c is an e ...

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Standard XII

Mathematics

Directrix of Ellipse

A curve is re...

Question

A curve is represented by c $\equiv$ $3 (x - 3 y + 2)^{2} + 2 (3 x + y - 1)^{2} = 180$ . Eccentricity of the curve is

\frac{1}{3}

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\frac{1}{\sqrt{3}}

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\frac{2}{3}

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\frac{2}{\sqrt{5}}

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Solution

The correct option is B $\frac{1}{\sqrt{3}}$
$3 (x - 3 y + 3)^{2} + 2 (3 x + y - 1)^{2} = 180 \Rightarrow \frac{(x - 3 y + 3)^{2}}{60} + \frac{(3 x + y - 1)}{90} = 1 \Rightarrow {(\frac{x - 3 y + 3}{\sqrt{1 + 3^{2}} \sqrt{6}})}^{2} + {(\frac{3 x + y - 1}{\sqrt{1 + 3^{2}} \times 3})}^{2} = 1$
$∴$ c is an ellipse whose lengths of axes are 6 and $2 \sqrt{6}$
$∴$ Ecentricity $= \frac{1}{\sqrt{3}}$

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A curve is represented by c ≡ 3(x–3y+2)2+2(3x+y–1)2=180. Eccentricity of the curve is

The correct option is B 1√33(x−3y+3)2+2(3x+y−1)2=180⇒ (x−3y+3)260+(3x+y−1)90=1⇒ (x−3y+3√1+32√6)2+(3x+y−1√1+32×3)2=1 ∴ c is an ellipse whose lengths of axes are 6 and 2√6 ∴ Ecentricity =1√3

A curve is represented by c $\equiv$ $3 (x - 3 y + 2)^{2} + 2 (3 x + y - 1)^{2} = 180$ . Eccentricity of the curve is