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The eccentricity of the conic 4(2 y-x-3)^2-9(2 x+y-1)^2=80 is 

Option: 1

\frac{3}{13}


Option: 2

\frac{\sqrt{13}}{3}


Option: 3

\sqrt{13}


Option: 4

3


Answers (1)

best_answer

(b) Here 2 y-x-3=0 \: \text{and} \: 2 x+y-1=0 are perpendicular to each other.
Therefore, the equation of the conic can be written as

\begin{aligned} & 4 \times 5\left[\frac{2 y-x-3}{\sqrt{2^2+1^2}}\right]^2-9 \times 5\left[\frac{2 x+y-1}{\sqrt{2^2+1^2}}\right]^2=80 \\ \\\Rightarrow & 4\left[\frac{2 y-x-3}{\sqrt{5}}\right]^2-9\left[\frac{2 x+y-1}{\sqrt{5}}\right]^2=16 \end{aligned}

On putting \frac{2 y-x-3}{\sqrt{5}}=X \text { and } \frac{2 x+y-1}{\sqrt{5}}=Y 

The given equation can be written as 

                     4 X^2-9 Y^2=16 \Rightarrow \frac{X^2}{4}-\frac{Y^2}{\left(\frac{4}{3}\right)^2}=1

which represents a hyperbola.

\text { Here, } a^2=4 \text { and } b^2=\frac{16}{9}

\begin{array}{ll} \therefore & b^2=a^2\left(e^2-1\right) \Rightarrow \frac{16}{9}=4\left(e^2-1\right) \\ \\\Rightarrow & e^2-1=\frac{4}{9} \Rightarrow e^2=1+\frac{4}{9}=\frac{13}{9} \\ \\\therefore & e=\frac{\sqrt{13}}{3} \end{array}

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Pankaj

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