Question

find the eccentricity of the conic x^2 -2x -4y^2=0​

Answer 1

Step-by-step explanation:

We can find the exact value of the eccentricity of these two conic shapes by using their equations. The eccentricity of an ellipse (x - h)2 / a2 + (y - k)2 / b2 = 1 will always be between 0 and 1 and can be calculated using the following formulas: When a > b, we use e = √(a2 - b2) / a.

Answer 2

Given:

$x^2 - 2x-4y^2 = 0$

To prove:

Eccentricity of the conic.

Solution:

Formula:

$\bold{\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1 } \\\\\bold{Eccentricity = \frac{\sqrt{a^2+b^2}}{a}}$

$\Rightarrow x^2-2x-4y^2 =0$

add one on both side of the above equation.

$\Rightarrow x^2-2x-4y^2+1 =0+1\\\\\Rightarrow x^2-2x+1 -4y^2 =1\\\\\Rightarrow (x-1)^2 -4y^2 =1$

$\Rightarrow (x-1)^2 -\frac{y^2}{\frac{1}{4}} =1\\\\\Rightarrow \frac{(x-1)^2}{(1)^2} -\frac{y^2}{(\frac{1}{2})^2} =1$

compare the value with above formula:

$\bold{\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}=1 }$

$a = 1\\b=\frac{1}{2}$

Formula:

$\Rightarrow eccentricity= \frac{\sqrt{a^2+b^2}}{a}\\\\\Rightarrow eccentricity= \frac{\sqrt{1^2+(\frac{1}{2})^2}}{1}\\\\\Rightarrow eccentricity= \frac{\sqrt{1+\frac{1}{4}}}{1}\\\\\Rightarrow eccentricity= \sqrt{\frac{5}{2^2}}\\\\\Rightarrow eccentricity= \frac{\sqrt{5}}{2}$

The Eccentricity of conic is $\bold{ = \frac{\sqrt{5}}{2}}$