
Find the area bounded by the ellipse \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] and the ordinates $x = 0$and $x = ae$, where ${b^2} = {a^2}\left( {1 - {e^2}} \right)$ and $e < 1$.
Answer
606.9k+ views
Hint: Simplify the given ellipse equation and integrate within the given ordinate limits to find the area.
An ellipse of the form \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] will meet the X-axis at (a, 0) and the Y-axis at (0, b). Let these points be P (a,0) and Q (0, b). It is symmetrical about the axes.
The ordinates given are $x = 0$and $x = ae$ which will be parallel to the Y-axis as shown in the figure.
The shaded area is the area bounded by the ellipse and the given ordinates.
Required area = Area of the shaded region
= $2 \times $Area of QOCD
=$2 \times \int_0^{ae} y dx$ …(1)
The given equation is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]. Let us find the value of y from this equation and substitute in equation (1).
$\begin{gathered}
\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1 \\
\begin{array}{*{20}{l}}
\begin{gathered}
\dfrac{{{y^2}}}{{{b^2}}} = 1 - \dfrac{{{x^2}}}{{{a^2}}} \\
\dfrac{{{y^2}}}{{{b^2}}} = \dfrac{{{a^2} - {x^2}}}{{{a^2}}} \\
\end{gathered} \\
{{y^2} = \dfrac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right)} \\
{y = \pm \sqrt {\dfrac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right)} } \\
{y = \pm \dfrac{b}{a}\sqrt {\left( {{a^2} - {x^2}} \right)} }
\end{array} \\
\end{gathered} $
Since, the area in equation (1) which is the area of QOCD is in the 1st quadrant. Hence, the value of y will be positive.
Hence, $y = \dfrac{b}{a}\sqrt {\left( {{a^2} - {x^2}} \right)} $ …(2)
Substituting (2) in (1),
Required area =$2 \times \int_0^{ae} y dx$
\[\begin{array}{*{20}{l}}
{ = 2\mathop \smallint \limits_0^{ae} \dfrac{b}{a}\sqrt {{a^2} - {x^2}} dx} \\
{ = \dfrac{{2b}}{a}\mathop \smallint \limits_0^{ae} \sqrt {{a^2} - {x^2}} dx}
\end{array}\] (Since a and b are constants)
We know that, \[\mathop \smallint \nolimits^ \sqrt {{a^2} - {x^2}} dx = \dfrac{{x\sqrt {{a^2} - {x^2}} }}{2} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}(\dfrac{x}{a}) + c\]
Using this in the previous step, we get
Required area = \[\dfrac{{2b}}{a}[\dfrac{1}{2}x\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a}]_0^{ae}\] \[\begin{gathered}
\begin{array}{*{20}{l}}
{ = \dfrac{{2b}}{a}[(\dfrac{{ae}}{2}\sqrt {{a^2} - {{(ae)}^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{{ae}}{a}) - (\dfrac{0}{2}\sqrt {{a^2} - 0} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(\dfrac{0}{a}))]} \\
{ = \dfrac{{2b}}{a}[\dfrac{{ae}}{2}\sqrt {{a^2} - {a^2}{e^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(e) - 0 - \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(0)]} \\
{ = \dfrac{{2b}}{a}[\dfrac{{ae}}{2} \cdot a\sqrt {1 - {e^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}e - 0]}
\end{array} \\
\begin{array}{*{20}{l}}
{ = \dfrac{{2b}}{a}[\dfrac{{{a^2}e}}{2}\sqrt {1 - {e^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}e]} \\
{ = \dfrac{{2b}}{a}(\dfrac{{{a^2}}}{2})[e\sqrt {1 - {e^2}} + {{\sin }^{ - 1}}e]} \\
{ = ab[e\sqrt {1 - {e^2}} + {{\sin }^{ - 1}}e]}
\end{array} \\
\end{gathered} \]
Required Area bounded by the ellipse \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] and the ordinates $x = 0$and $x = ae$
\[ = ab[e\sqrt {1 - {e^2}} + {\sin ^{ - 1}}e]\]
Note: The required area can also be found by integrating the entire shaded area QOABCD instead of finding $2 \times $Area of QOCD. It would be a little lengthier and more unnecessary because the given ellipse is symmetrical about the origin.
An ellipse of the form \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] will meet the X-axis at (a, 0) and the Y-axis at (0, b). Let these points be P (a,0) and Q (0, b). It is symmetrical about the axes.
The ordinates given are $x = 0$and $x = ae$ which will be parallel to the Y-axis as shown in the figure.
The shaded area is the area bounded by the ellipse and the given ordinates.
Required area = Area of the shaded region
= $2 \times $Area of QOCD
=$2 \times \int_0^{ae} y dx$ …(1)
The given equation is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]. Let us find the value of y from this equation and substitute in equation (1).
$\begin{gathered}
\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1 \\
\begin{array}{*{20}{l}}
\begin{gathered}
\dfrac{{{y^2}}}{{{b^2}}} = 1 - \dfrac{{{x^2}}}{{{a^2}}} \\
\dfrac{{{y^2}}}{{{b^2}}} = \dfrac{{{a^2} - {x^2}}}{{{a^2}}} \\
\end{gathered} \\
{{y^2} = \dfrac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right)} \\
{y = \pm \sqrt {\dfrac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right)} } \\
{y = \pm \dfrac{b}{a}\sqrt {\left( {{a^2} - {x^2}} \right)} }
\end{array} \\
\end{gathered} $
Since, the area in equation (1) which is the area of QOCD is in the 1st quadrant. Hence, the value of y will be positive.
Hence, $y = \dfrac{b}{a}\sqrt {\left( {{a^2} - {x^2}} \right)} $ …(2)
Substituting (2) in (1),
Required area =$2 \times \int_0^{ae} y dx$
\[\begin{array}{*{20}{l}}
{ = 2\mathop \smallint \limits_0^{ae} \dfrac{b}{a}\sqrt {{a^2} - {x^2}} dx} \\
{ = \dfrac{{2b}}{a}\mathop \smallint \limits_0^{ae} \sqrt {{a^2} - {x^2}} dx}
\end{array}\] (Since a and b are constants)
We know that, \[\mathop \smallint \nolimits^ \sqrt {{a^2} - {x^2}} dx = \dfrac{{x\sqrt {{a^2} - {x^2}} }}{2} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}(\dfrac{x}{a}) + c\]
Using this in the previous step, we get
Required area = \[\dfrac{{2b}}{a}[\dfrac{1}{2}x\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a}]_0^{ae}\] \[\begin{gathered}
\begin{array}{*{20}{l}}
{ = \dfrac{{2b}}{a}[(\dfrac{{ae}}{2}\sqrt {{a^2} - {{(ae)}^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{{ae}}{a}) - (\dfrac{0}{2}\sqrt {{a^2} - 0} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(\dfrac{0}{a}))]} \\
{ = \dfrac{{2b}}{a}[\dfrac{{ae}}{2}\sqrt {{a^2} - {a^2}{e^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(e) - 0 - \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(0)]} \\
{ = \dfrac{{2b}}{a}[\dfrac{{ae}}{2} \cdot a\sqrt {1 - {e^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}e - 0]}
\end{array} \\
\begin{array}{*{20}{l}}
{ = \dfrac{{2b}}{a}[\dfrac{{{a^2}e}}{2}\sqrt {1 - {e^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}e]} \\
{ = \dfrac{{2b}}{a}(\dfrac{{{a^2}}}{2})[e\sqrt {1 - {e^2}} + {{\sin }^{ - 1}}e]} \\
{ = ab[e\sqrt {1 - {e^2}} + {{\sin }^{ - 1}}e]}
\end{array} \\
\end{gathered} \]
Required Area bounded by the ellipse \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] and the ordinates $x = 0$and $x = ae$
\[ = ab[e\sqrt {1 - {e^2}} + {\sin ^{ - 1}}e]\]
Note: The required area can also be found by integrating the entire shaded area QOABCD instead of finding $2 \times $Area of QOCD. It would be a little lengthier and more unnecessary because the given ellipse is symmetrical about the origin.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

