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This is the second of our many quizzes on Eccentricity of conics. Click on this to access more such competitive quizzes on 2D Coordinate Geometry.

Eccentricity of conics Quiz 2

mathematics

Prateek has a conic section with its eccentricity measured as $\text{0<}e\text{<1}$. What is the conic section Prateek has?

 The general value of eccentricity is given in the question, calculate which figure the range is closest.

The formula to find out the eccentricity of any conic section is defined as Eccentricity, $e=\dfrac{c}{a}$ Where c = distance from the centre to the focus, a = distance from the centre to the vertex The eccentricity of a circle is zero. The eccentricity of an ellipse which is not a circle is greater than zero but less than 1. The eccentricity of a parabola is 1. The eccentricity of a hyperbola is greater than 1.

 Vrushali has a hyperbola where the distance of the semi major axis is 3 and the distance of the semi minor axis is 2. What would be the eccentricity?  

 The eccentricity of hyperbola is given by $e=\dfrac{c}{a}$where $c=\sqrt{{{a}^{2}}+{{b}^{2}}}$

The eccentricity of hyperbola is given by $e=\dfrac{c}{a}$Where $c=\sqrt{{{a}^{2}}+{{b}^{2}}}$ Finding the value of c $\Rightarrow c=\sqrt{{{3}^{2}}+{{2}^{2}}}$ $\Rightarrow c=\sqrt{9+4}=\sqrt{13}$ $e=\dfrac{\sqrt{13}}{3}$

 For the given equation of ellipse $\dfrac{{{x}^{2}}}{144}+\dfrac{{{y}^{2}}}{81}=1$. Find the ratio of the distances from any point to a focus and the corresponding directrix.

 The value of the denominator with ${{x}^{2}}$  as a numerator is greater. Hence, the equation is of the following form:$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$.

In the equation, the value of the denominator of the first fraction with ${{x}^{2}}$ as a numerator is greater than the second one. Hence, the equation is of the following form: $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ It shows that the ellipse has a horizontal major axis. Here ${{a}^{2}}=144$ so $a=12$ Similarly ${{b}^{2}}=81$ so $b=9$ The formula for computing the eccentricity of the ellipse is given below: $e=\dfrac{c}{a}$ Where $c=\sqrt{{{a}^{2}}-{{b}^{2}}}$ Replacing the values of a and b \[c=\sqrt{{{12}^{2}}-{{9}^{2}}}=\sqrt{144-81}=\sqrt{144-81}=\sqrt{63}\] Finding the eccentricity $e=\dfrac{\sqrt{63}}{12}$

 In a radio system, rectangular hyperbola is taken into use. The length of its major axis is “x” and the minor axis is “y”. Find its eccentricity in terms of x and y  

 Since from the equation we can see that it is a rectangular hyperbolais a rectangular hyperbola, the length of the major axis will be equal to the length of the minor axis. 

From the equation we can see that it is a rectangular hyperbola $\text{length of major axis = length of minor axis}$ $\therefore \text{ x = y as per the question}$ The eccentricity of hyperbola is given by $e=\dfrac{c}{a}$Where $c=\sqrt{{{a}^{2}}+{{b}^{2}}}$ Finding the value of c $c=\sqrt{{{x}^{2}}+{{x}^{2}}}=x\sqrt{2}$ Calculating eccentricity $e=\dfrac{x\sqrt{2}}{x}=\sqrt{2}$ The value is independent of x and y and is equal to $\sqrt{2}$.

 Manish wants to make a mural shaped as ellipse, he plots the ellipse with the equation $20{{x}^{2}}+5{{y}^{2}}=80$. What would be the eccentricity calculated by him?

 The formula for computing the eccentricity of the ellipse is given below:$e=\dfrac{c}{a}$ Where $c=\sqrt{{{b}^{2}}-{{a}^{2}}}$.

Converting the equation into standard form of ellipse \[\dfrac{20}{80}{{x}^{2}}+\dfrac{5}{80}{{y}^{2}}=1\] \[\Rightarrow \dfrac{1}{4}{{x}^{2}}+\dfrac{1}{16}{{y}^{2}}=1\] Since the denominator of ${{y}^{2}}$ is greater than the denominator of ${{x}^{2}}$, therefore we can say that this ellipse has a vertical major axis. We get the values of a and be as 2 and 4 respectively The formula for computing the eccentricity of the ellipse is given below: $e=\dfrac{c}{a}$ Where $c=\sqrt{{{b}^{2}}-{{a}^{2}}}$ Replacing the values of a and b \[c=\sqrt{{{4}^{2}}-{{2}^{2}}}=\sqrt{16-4}=\sqrt{12}\] Finding the eccentricity \[e=\dfrac{\sqrt{12}}{4}\]