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Ellipse properties Quiz 2

mathematics

 Which of the following ellipses has coordinates of foci $\left( \pm \dfrac{2}{\sqrt{105}},0 \right)$

 Compare the given equation of ellipse with the general equation of ellipse and find ${{a}^{2}}$ and ${{b}^{2}}$, Then find c.

  $  \text{Consider the equation of ellipse }15{{x}^{2}}+35{{y}^{2}}=1 \\  \Rightarrow \dfrac{{{x}^{2}}}{\left( \dfrac{1}{15} \right)}+\dfrac{{{y}^{2}}}{\left( \dfrac{1}{35} \right)}=1.........(i) \\  \text{Consider the general equation of ellipse} \\  \Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1..........(ii) \\  \text{Compare equation }\left( \text{i} \right)\text{ and }\left( \text{ii} \right) \\  \Rightarrow {{a}^{2}}=\dfrac{1}{15}\text{ and }{{b}^{2}}=\dfrac{1}{35} \\  \text{Now, }c=\sqrt{{{a}^{2}}-{{b}^{2}}}=\sqrt{\dfrac{1}{15}-\dfrac{1}{35}}=\sqrt{\dfrac{4}{105}}=\pm \dfrac{2}{\sqrt{105}} \\  \text{Foci of ellipse}=\left( \pm c,0 \right)=\left( \pm \dfrac{2}{\sqrt{105}},0 \right) \\ $

 At which points on the $x-$axis, the ellipse $17\left( {{x}^{2}}+3{{y}^{2}} \right)=9$ intersects. Also, write the equation of the ellipse in standard form.

 To find the point on the $x-\text{axis},$ put $y=0$ in the given equation to write the equation in standard form of ellipse. Make coefficients of ${{x}^{2}}$ and ${{y}^{2}}$ as $1.$ 

  $  \text{Given equation of ellipse is} \\  17\left( {{x}^{2}}+3{{y}^{2}} \right)=9..........(i) \\  \text{Put }y=0\text{ in equation }\left( \text{i} \right) \\  \Rightarrow 17{{x}^{2}}=9 \\  \Rightarrow {{x}^{2}}=\dfrac{9}{17} \\  \Rightarrow x\pm \dfrac{3}{\sqrt{17}} \\  \text{Ellipse intersects the x-axis at the point }\left( \pm \dfrac{3}{\sqrt{17}},0 \right). \\  \text{Equation }\left( \text{i} \right)\text{ can also be written as,} \\  \Rightarrow {{x}^{2}}+3{{y}^{2}}=\dfrac{9}{17} \\  \text{Divide both sides by }\dfrac{9}{17} \\  \Rightarrow \dfrac{{{x}^{2}}}{\left( \dfrac{9}{17} \right)}+\dfrac{3{{y}^{2}}}{\left( \dfrac{9}{17} \right)}=1 \\  \Rightarrow \dfrac{{{x}^{2}}}{\left( \dfrac{9}{17} \right)}+\dfrac{{{y}^{2}}}{\dfrac{9}{17\times 3}}=1 \\  \Rightarrow \dfrac{{{x}^{2}}}{\left( \dfrac{9}{17} \right)}+\dfrac{{{y}^{2}}}{\left( \dfrac{3}{17} \right)}=1 \\ $

 Write the value of length of latus rectum of ellipse $133{{x}^{2}}+152{{y}^{2}}=112.$

 First convert the given equation in general form of ellipse and find ${{a}^{2}}$ and ${{b}^{2}}.$ Then find the length of the latus rectum by putting values of ${{a}^{2}}$ and ${{b}^{2}}$ in the formula to find the length of latus rectum. 

$  \text{Given equation of ellipse }133{{x}^{2}}+152{{y}^{2}}=112 \\  \text{Divide both sides by }112 \\  \Rightarrow \dfrac{133{{x}^{2}}}{112}+\dfrac{152{{y}^{2}}}{112}=1 \\  \Rightarrow \dfrac{{{x}^{2}}}{\dfrac{112}{133}}+\dfrac{{{y}^{2}}}{\dfrac{112}{152}}=1.......(i) \\  \text{Consider the general equation of ellipse} \\  \Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1..........(ii) \\  \text{From equations }\left( \text{i} \right)\text{ and }\left( \text{ii} \right)\text{, we get} \\  \Rightarrow {{a}^{2}}=\dfrac{112}{133}\text{ and }{{b}^{2}}=\dfrac{112}{152} \\  \text{Length of latus rectum}=\dfrac{2{{b}^{2}}}{a} \\  \Rightarrow \dfrac{2\times \dfrac{112}{152}}{\sqrt{\dfrac{112}{133}}} \\  =\dfrac{28}{19}\times \dfrac{\sqrt{133}}{\sqrt{112}} \\  =\dfrac{7\sqrt{19}}{19} \\ $

 Write the eccentricity of the ellipse $5{{x}^{2}}+9{{y}^{2}}=29.$

 Convert the given equation into a general equation of ellipse and find ${{a}^{2}}$ and ${{b}^{2}}.$ Then find $c$. Finally, find eccentricity by putting values of $c$ and $a$ in the formula to find the eccentricity of the ellipse. 

 $  \text{The given equation of ellipse is,} \\  \Rightarrow 5{{x}^{2}}+9{{y}^{2}}=29 \\  \text{Divide both sides by }29 \\  \dfrac{5{{x}^{2}}}{29}+\dfrac{9{{y}^{2}}}{29}=1 \\  \Rightarrow \dfrac{{{x}^{2}}}{\left( \dfrac{29}{5} \right)}+\dfrac{{{y}^{2}}}{\left( \dfrac{29}{9} \right)}=1........(i) \\  \text{Consider the general equation of ellipse} \\  \Rightarrow \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1.........(ii) \\  \text{Compare equation }\left( \text{i} \right)\text{ and }\left( \text{ii} \right) \\  {{a}^{2}}=\dfrac{29}{5}\text{ and }{{b}^{2}}=\dfrac{29}{9} \\  \text{Now, }c=\sqrt{{{a}^{2}}-{{b}^{2}}}=\sqrt{\dfrac{29}{5}-\dfrac{29}{9}}=\dfrac{2}{3}\sqrt{\dfrac{29}{5}} \\  \text{Consider eccentricity of ellipse,} \\  \Rightarrow e=\dfrac{c}{a} \\  \Rightarrow e=\dfrac{\dfrac{2}{3}\sqrt{\dfrac{29}{5}}}{\sqrt{\dfrac{29}{5}}} =\dfrac{2}{3} \\ $

 Write the points on the ellipse $6{{x}^{2}}+7{{y}^{2}}=1$ where the tangent is parallel to $x-$axis.

 Find $\dfrac{dy}{dx}$ and equate it to zero to find the $x-$coordinate of the required point. Put this value of $x-$coordinate in the given equation to find $y-$coordinate.

 $  \text{The given equation of ellipse is,} \\  \Rightarrow 6{{x}^{2}}+7{{y}^{2}}=1.........(i) \\  \text{Differentiate both sides with respect to} \\  \Rightarrow 12x+14y\dfrac{dy}{dx}=0 \\  \Rightarrow 14y\dfrac{dy}{dx}=-12x \\  \Rightarrow \dfrac{dy}{dx}=-\dfrac{6x}{7y} \\  \text{For tangent parallel to }x-\text{axis,}\dfrac{dy}{dx}=0 \\  \Rightarrow -\dfrac{6x}{7y}=0 \\  \Rightarrow x=0 \\  \text{Put }\!\!~\!\!\text{  }x=0\text{ in equation }\left( \text{i} \right) \\  \Rightarrow 7{{y}^{2}}=1 \\  \Rightarrow y=\pm \dfrac{1}{\sqrt{17}} \\  \therefore \text{The required point is }\left( 0,\pm \dfrac{1}{\sqrt{17}} \right) \\ $