If the length of the semi-major axis of an ellipse is 68 and the eccentricity is $ \dfrac{1}{2} $ and if the area of the rectangle formed by joining the vertices of the latus rectum of the ellipse is $ \Delta $ . Then find the value of $ \dfrac{\Delta }{867} $ .
Answer
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Hint: We first need to find the general equation of an ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ . Based on the equation we find the value of the length of the semi major axis of that ellipse and the eccentricity. We place the values in the equation $ {{b}^{2}}={{a}^{2}}\left( 1-{{e}^{2}} \right) $ to find the value of b. We use those values and the concept of area to find the value of $ \Delta $ and $ \dfrac{\Delta }{867} $ .
Complete step by step answer:
For a general equation of an ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ , the length of the semi major axis of an ellipse is 68 and the eccentricity is $ \dfrac{1}{2} $ .
This means $ a=68 $ and $ e=\dfrac{1}{2} $ .
We have the formula to find the value of b which is $ {{b}^{2}}={{a}^{2}}\left( 1-{{e}^{2}} \right) $ . Placing the values, we get $ {{b}^{2}}={{68}^{2}}\left( 1-\dfrac{1}{4} \right)=3\times {{34}^{2}} $ .
We know that area of the rectangle formed by joining latus rectum is equal to the multiplication of the length equal to $ \dfrac{2{{b}^{2}}}{a} $ and breadth, the distance between the latus rectum equal to 2ae.
We have been given that the area formed by joining the vertices of the latus rectum of the ellipse is $ \Delta $ . So, $ \Delta =\dfrac{2{{b}^{2}}}{a}.2ae=\dfrac{4\times 3\times {{34}^{2}}}{2}=6936 $ .
We need to find the value of $ \dfrac{\Delta }{867} $ . Therefore, $ \dfrac{\Delta }{867}=\dfrac{6936}{867}=8 $ .
The value of $ \dfrac{\Delta }{867} $ is 8.
Note:
Considering the general equation of ellipse as $ \dfrac{{{\left( x-\alpha \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-\beta \right)}^{2}}}{{{b}^{2}}}=1 $ will not complicate the solution as we are dealing with area and their major and minor axes. Having a different centre other than $ \left( 0,0 \right) $ only changes the position of the ellipse, not the area.
Complete step by step answer:
For a general equation of an ellipse $ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $ , the length of the semi major axis of an ellipse is 68 and the eccentricity is $ \dfrac{1}{2} $ .
This means $ a=68 $ and $ e=\dfrac{1}{2} $ .
We have the formula to find the value of b which is $ {{b}^{2}}={{a}^{2}}\left( 1-{{e}^{2}} \right) $ . Placing the values, we get $ {{b}^{2}}={{68}^{2}}\left( 1-\dfrac{1}{4} \right)=3\times {{34}^{2}} $ .
We know that area of the rectangle formed by joining latus rectum is equal to the multiplication of the length equal to $ \dfrac{2{{b}^{2}}}{a} $ and breadth, the distance between the latus rectum equal to 2ae.
We have been given that the area formed by joining the vertices of the latus rectum of the ellipse is $ \Delta $ . So, $ \Delta =\dfrac{2{{b}^{2}}}{a}.2ae=\dfrac{4\times 3\times {{34}^{2}}}{2}=6936 $ .
We need to find the value of $ \dfrac{\Delta }{867} $ . Therefore, $ \dfrac{\Delta }{867}=\dfrac{6936}{867}=8 $ .
The value of $ \dfrac{\Delta }{867} $ is 8.
Note:
Considering the general equation of ellipse as $ \dfrac{{{\left( x-\alpha \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-\beta \right)}^{2}}}{{{b}^{2}}}=1 $ will not complicate the solution as we are dealing with area and their major and minor axes. Having a different centre other than $ \left( 0,0 \right) $ only changes the position of the ellipse, not the area.
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