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This is one of our many quizzes on ellipse. Click on this to access more such competitive quizzes on ellipse.

Ellipse Quiz 2

mathematics

 Consider a semi-ellipse which is shaped into an arc. Let us suppose that the height and wideness are 5 meters and 7 meters respectively. Figure out the height of the arc at a point of 2 meters from any of its ends.

 First put the values of ‘a’ and ‘b’ in the standard elliptical equation to obtain the equation$\dfrac{4{{x}^{2}}}{49}+\dfrac{{{y}^{2}}}{25}=1$ of ellipse and then use substitution.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c8d2194d-92fb-44f6-ac09-977b2bef59293102327929883669234.png"/>$\text{Since, }a=OJ=\dfrac{7}{2},b=5 \\ \Rightarrow \dfrac{{{x}^{2}}}{{{\left( \dfrac{7}{2} \right)}^{2}}}+\dfrac{{{y}^{2}}}{{{\left( 5 \right)}^{2}}}=1 \\ \Rightarrow \dfrac{{{x}^{2}}}{\dfrac{49}{4}}+\dfrac{{{y}^{2}}}{25}=1 \\ \Rightarrow \dfrac{4{{x}^{2}}}{49}+\dfrac{{{y}^{2}}}{25}=1 \\$ To find the value of the correct height of the same arc at a point of $2m$ from any one of its ends. $ \\ \text{L =}\left( ON,LN \right) \\ \Rightarrow \left( OJ-NJ,LN \right) \\ \Rightarrow \left( \dfrac{7}{2}-2,LN \right)\,\,\,\because NJ=2m \\ \Rightarrow \left( \dfrac{3}{2},LN \right) \\ \Rightarrow LN=3.2 \\ $

 Like Earth, Planet Pamj also travels in an elliptical orbit, $3{{x}^{2}}+2x+{{y}^{2}}=1$, find the center for its path.

 First write the equation in the standard form of an ellipse,  \[\dfrac{{{\left(x-b\right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left(y-c\right)}^{2}}}{{{d}^{2}}}=1\], where $\left(b,c\right)$ is the center of the elliptical path.

  $\text{Here, }3{{x}^{2}}+2x+{{y}^{2}}=1 \\ 3\left({{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}\right)+\left({{y}^{2}}\right)=1+\dfrac{1}{3} \\ \dfrac{9}{4}{{\left(x+\dfrac{1}{3}\right)}^{2}}+\dfrac{3}{4}{{y}^{2}}=1 \\ \dfrac{{{\left(x+\dfrac{1}{3}\right)}^{2}}}{{{\left(\dfrac{2}{3}\right)}^{2}}}+\dfrac{{{y}^{2}}}{{{\left(\dfrac{2}{\sqrt{3}}\right)}^{2}}}=1 \\ \text{Center }= \left(-\dfrac{1}{3},0\right) $

 Which of the following equations of ellipse will have eccentricity $\dfrac{13.22}{16}$?

 Use eccentricity formula $e=\sqrt{1-\dfrac{{{a}^{2}}}{{{b}^{2}}}}$.

 $81{{x}^{2}}+256{{y}^{2}}=20736 \\ \text{Divide the equation by 20736.} \\  \Rightarrow \dfrac{{{x}^{2}}}{{{16}^{2}}}+\dfrac{{{y}^{2}}}{{{9}^{2}}}=1 \\ \Rightarrow a=9,b=16 \\ \text{Since,}e=\sqrt{1-\dfrac{{{9}^{2}}}{{{16}^{2}}}}=\dfrac{13.22}{16} \\ $ 

 If foci are $\left(-1,5\right),\left(1,5\right)$ along with the semi and major axis $6$ , then which of the following will be the correct equation of this type of ellipse?

 Find value of a first by eccentricity formula and after that use distance formula to get the required equation. 

 Since, the distance between the two foci of the ellipse is always written in the form of (-ae, 0), (ae, 0) and since, given foci are (-1, 5), (1, 5) So, the distance of AB is 2 ae due to the presence of – 1 on the left and + 1 on the right side of the ellipse. $\\ $ Since ${b = 6, ae = 1}$ So, $\\ \Rightarrow 1-{{e}^{2}}=\dfrac{{{b}^{2}}}{{{a}^{2}}} \\ \Rightarrow {{a}^{2}}-{{\left( ea \right)}^{2}}={{b}^{2}} \\ \Rightarrow a=\sqrt{37} \\ \text{W.K.T, }A=\left(-1,5\right),B=\left(1,5\right),P=\left( x,y \right) \\$ Since, $AP+BP=2a \\ \Rightarrow \sqrt{{{\left( x+1 \right)}^{2}}+{{\left( y-5 \right)}^{2}}}+\sqrt{{{\left( x-1 \right)}^{2}}+{{\left( y-5 \right)}^{2}}}=2\sqrt{37} \\ \Rightarrow {{x}^{2}}+{{y}^{2}}-10y+26=74 \\ $ 

 What will be the value of vertices if the equation of the ellipse is ${{x}^{2}}+3{{y}^{2}}-6x-12y+20=0$?

 First bring the given equation to the standard form of equation then use eccentricity formula $e=\sqrt{1-\dfrac{{{a}^{2}}}{{{b}^{2}}}}$.

 ${{x}^{2}}+3{{y}^{2}}-6x-12y+20=0 \\ \Rightarrow {{x}^{2}}+3{{y}^{2}}-6x-12y+21=1 \\ \Rightarrow {{x}^{2}}+9-6x+3\left( {{y}^{2}}+4-4y \right)=1 \\ \Rightarrow {{\left( x-3 \right)}^{2}}+3{{\left( y-2 \right)}^{2}}=1 \\ \Rightarrow \dfrac{{{\left( x-3 \right)}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}}+\dfrac{{{\left( y-2 \right)}^{2}}}{{{1}^{2}}}=\dfrac{1}{3} \\ \Rightarrow \dfrac{{{X}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}}+\dfrac{{{Y}^{2}}}{{{1}^{2}}}=\dfrac{1}{3} \\ \text{Since, }e=\sqrt{1-\dfrac{{{a}^{2}}}{{{b}^{2}}}}\,\text{where}\,a=1,b=\sqrt{3} \\ \Rightarrow e=\sqrt{1-\dfrac{{{1}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}}}=\sqrt{\dfrac{2}{3}} \\ $ 