Find the area of the region bounded by the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{5}=1$ between the two latus rectums.

> [!TLDR] > Locate the latus rectums at $x=\pm c$, then integrate $y=\dfrac{b}{a}\sqrt{a^2-x^2}$ between them using symmetry. > > 1. **Identify $a$, $b$, $c$.** From $\dfrac{x^2}{9}+\dfrac{y^2}{5}=1$: $a^2=9\Rightarrow a=3$; $b^2=5\Rightarrow b=\sqrt5$. > > $c^2=a^2-b^2=9-5=4\Rightarrow c=2$ > > The latus rectums are the vertical chords through the foci $x=\pm c=\pm2$. > > 2. **Set up the area integral.** The ellipse is symmetric about both axes, so the area between $x=-2$ and $x=2$ is > > $A=4\displaystyle\int_0^{2} y\,dx$, where $y=\dfrac{b}{a}\sqrt{a^2-x^2}=\dfrac{\sqrt5}{3}\sqrt{9-x^2}$ (taking the upper half and doubling for the lower half, then doubling again for both sides of the $y$-axis). > > $A=4\cdot\dfrac{\sqrt5}{3}\displaystyle\int_0^{2}\sqrt{9-x^2}\,dx$ > > 3. **Evaluate the integral** using $\displaystyle\int\sqrt{p^2-x^2}\,dx=\dfrac{x}{2}\sqrt{p^2-x^2}+\dfrac{p^2}{2}\sin^{-1}\dfrac{x}{p}+C$ with $p=3$: > > $\displaystyle\int_0^2\sqrt{9-x^2}\,dx=\left[\dfrac{x}{2}\sqrt{9-x^2}+\dfrac{9}{2}\sin^{-1}\dfrac{x}{3}\right]_0^2$ > > At $x=2$: $\dfrac{2}{2}\sqrt{9-4}+\dfrac{9}{2}\sin^{-1}\dfrac{2}{3}=\sqrt5+\dfrac{9}{2}\sin^{-1}\dfrac{2}{3}$ > > At $x=0$: $0$ > …

Find the area of the region bounded by the ellipse (x²)/(9)+(y²)/(5)=1 between the two latus rectums.

Question 57 of 96

\dfrac{x^2}{9}+\dfrac{y^2}{5}=1

Puducherry TnboardTamil Nadu HSC (DGE) Board 2016Subjective· 10mImportance★★★★★

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Concept understanding — Area of a Bounded Plane Region by Integration

Building on the geometric meaning of $\int_a^b f(x)\,dx$ (Remarks under the limit-of-a-sum definition), this topic packages the vertical/horizontal-strip argument into ready-to-use area formulas.

Bounded by a curve, the $x$ -axis, and $x=a,\,x=b$ (vertical strips, height $=|y|$ , width $\Delta x$ ):

If $y=f(x)\ge0$ throughout $[a,b]$ (curve above the $x$ -axis): $A=\displaystyle\int_a^b y\,dx$ .
If $y=f(x)\le0$ throughout $[a,b]$ (curve below the $x$ -axis): $A=\displaystyle-\int_a^b y\,dx=\left|\int_a^b y\,dx\right|$ .
If $f$ changes sign on $[a,b]$ : split at the zeros $c_1,c_2,\ldots$ into subintervals of constant sign, apply the two rules above on each piece, and add the absolute values — $A=\left|\int_a^{c_1}f\right|+\left|\int_{c_1}^{c_2}f\right|+\cdots$ (never just $\int_a^b f\,dx$ , which can cancel positive against negative area).

Bounded by a curve, the $y$ -axis, and $y=c,\,y=d$ (horizontal strips, mirror image): $A=\displaystyle\int_c^d x\,dy$ if the curve lies to the right of the $y$ -axis ( $x\ge0$ ); $A=-\displaystyle\int_c^d x\,dy$ if to the left; split-and-add-absolute-values if it crosses.

Bounded between two curves. If $f(x)\ge g(x)$ on $[a,b]$ (an "upper" curve $U$ and "lower" curve $L$ ), the region between them and the ordinates $x=a,x=b$ has area

A=\int_a^b[f(x)-g(x)]\,dx=\int_a^b(y_U-y_L)\,dx.

The $y$ -axis mirror (curves $x=f(y)\ge x=g(y)$ , a "right" curve $R$ and "left" curve $L$ ) gives $A=\int_c^d(x_R-x_L)\,dy$ . …

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Locate the latus rectums at $x=\pm c$ , then integrate $y=\dfrac{b}{a}\sqrt{a^2-x^2}$ between them using symmetry.

Identify $a$ , $b$ , $c$ . From $\dfrac{x^2}{9}+\dfrac{y^2}{5}=1$ : $a^2=9\Rightarrow a=3$ ; $b^2=5\Rightarrow b=\sqrt5$ .

$c^2=a^2-b^2=9-5=4\Rightarrow c=2$

The latus rectums are the vertical chords through the foci $x=\pm c=\pm2$ .
Set up the area integral. The ellipse is symmetric about both axes, so the area between $x=-2$ and $x=2$ is

$A=4\displaystyle\int_0^{2} y\,dx$ , where $y=\dfrac{b}{a}\sqrt{a^2-x^2}=\dfrac{\sqrt5}{3}\sqrt{9-x^2}$ (taking the upper half and doubling for the lower half, then doubling again for both sides of the $y$ -axis).

$A=4\cdot\dfrac{\sqrt5}{3}\displaystyle\int_0^{2}\sqrt{9-x^2}\,dx$
Evaluate the integral using $\displaystyle\int\sqrt{p^2-x^2}\,dx=\dfrac{x}{2}\sqrt{p^2-x^2}+\dfrac{p^2}{2}\sin^{-1}\dfrac{x}{p}+C$ with $p=3$ :

$\displaystyle\int_0^2\sqrt{9-x^2}\,dx=\left[\dfrac{x}{2}\sqrt{9-x^2}+\dfrac{9}{2}\sin^{-1}\dfrac{x}{3}\right]_0^2$

At $x=2$ : $\dfrac{2}{2}\sqrt{9-4}+\dfrac{9}{2}\sin^{-1}\dfrac{2}{3}=\sqrt5+\dfrac{9}{2}\sin^{-1}\dfrac{2}{3}$

At $x=0$ : $0$

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Q.Find the area of the region bounded by the ellipse x29+y25=1\dfrac{x^2}{9}+\dfrac{y^2}{5}=19x2​+5y2​=1 between the two latus rectums.

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Q.Find the area of the region bounded by the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{5}=1$ between the two latus rectums.

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