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
385.2k+ 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
Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts
State Gay Lusaaccs law of gaseous volume class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

Which is the tallest animal on the earth A Giraffes class 9 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
