A racing track is build around an elliptical ground whose equation is given by 9x² + 16y² = 144. The width of

Question

9x^2 + 16y^2 = 144

(i) Express

y

as a function of

x

from the given equation of ellipse.

(ii) Integrate the function obtained in

(i) with respect to

x

(iii)

(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.

(OR)

(iii)

(b) Write the co-ordinates of the points P and Q where the outer edge of the track cuts

x

axis and

y

axis in first quadrant and find the area of the triangle formed by points P, O, Q using integration.

CBSECBSE Class XII Board 2026Subjective· 4mImportance★★★★★

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Part (a)Concept understanding — Area of Ellipse

Area of an Ellipse

A circle of radius $r$ has area $\pi r^2$ . An ellipse is a circle that has been stretched by different amounts along two perpendicular directions, so it is natural to expect its area to be a stretched version of $\pi r^2$ . The standard ellipse centred at the origin is

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,

where $a$ is the semi-major (or semi-minor) axis along $x$ and $b$ is the semi-axis along $y$ . The result we want is beautifully simple:

\text{Area of ellipse} = \pi a b

Notice that when $a = b = r$ the ellipse becomes a circle and $\pi ab$ collapses to $\pi r^2$ — a good sanity check.

Finding it by integration

Because the ellipse is symmetric about both axes, we compute the area of the piece in the first quadrant and multiply by $4$ . Solving the equation for the upper half gives

y = b\sqrt{1 - \frac{x^2}{a^2}} = \frac{b}{a}\sqrt{a^2 - x^2}.

As $x$ runs from $0$ to $a$ this traces the first-quadrant arc, so

\text{Area} = 4 \int_0^a \frac{b}{a}\sqrt{a^2 - x^2}\, dx.

The integral $\displaystyle\int_0^a \sqrt{a^2 - x^2}\, dx$ is the area of a quarter-circle of radius $a$ , which equals $\dfrac{\pi a^2}{4}$ . (You may also get it from the standard result $\int \sqrt{a^2 - x^2}\,dx = \tfrac{x}{2}\sqrt{a^2-x^2} + \tfrac{a^2}{2}\sin^{-1}\tfrac{x}{a} + C$ .) Therefore

\text{Area} = \frac{4b}{a} \cdot \frac{\pi a^2}{4} = \pi a b.

The intuition

The factor $\dfrac{b}{a}$ in front is exactly the vertical stretch that turns a circle of radius $a$ into this ellipse: it scales every height by $b/a$ , and scaling all heights scales the area by the same ratio. Multiplying the circle's area $\pi a^2$ by $b/a$ gives $\pi ab$ . …

Part (b)Concept understanding — Area Under Curve

Area Under a Curve

How do you measure the area of a region whose top edge is curved rather than a straight line? For rectangles and triangles we have formulas, but a shape bounded above by $y = f(x)$ has no simple side lengths to plug in. The definite integral is the tool built exactly for this.

The core idea: area as a limit of strips

Take the region under $y = f(x)$ (with $f(x) \ge 0$ ), above the $x$ -axis, between $x = a$ and $x = b$ . Slice it into many thin vertical strips. A strip at position $x$ with tiny width $dx$ is almost a rectangle of height $f(x)$ , so its area is about $f(x)\, dx$ . Add up all the strips and let their width shrink to zero: the sum becomes the definite integral

\text{Area} = \int_a^b f(x)\, dx.

This is why the integral is the area — it is the exact total of infinitely many infinitesimally thin rectangles.

How to compute it

Find an antiderivative $F(x)$ (so $F'(x) = f(x)$ ) and evaluate at the two limits — the Fundamental Theorem of Calculus:

\int_a^b f(x)\, dx = F(b) - F(a).

For example, the area under $y = x^2$ from $0$ to $1$ is $\left[\tfrac{x^3}{3}\right]_0^1 = \tfrac13$ .

Cases you must handle carefully

Watch out

The integral gives signed area. Where the curve dips below the $x$ -axis, $f(x) < 0$ and $\int f\,dx$ comes out negative. For the geometric (positive) area of such a stretch, integrate the absolute value or take the magnitude of that piece: $\displaystyle \text{Area} = \int_a^b |f(x)|\, dx$ .

If a curve crosses the $x$ -axis inside $[a,b]$ , split the integral at each crossing and add the sizes of the parts.

Area with respect to the y-axis

When the region is bounded by a curve $x = g(y)$ and the $y$ -axis between $y = c$ and $y = d$ , slice horizontally instead: …

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The ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$ , so $y=\frac34\sqrt{16-x^2}$ with $\int y\,dx=\frac38\!\left(x\sqrt{16-x^2}+16\sin^{-1}\frac x4\right)+C$ . (iii)(a) ground area $=12\pi\ \text{m}^2$ . (iii)(b) $P(7,0),\ Q(0,6)$ and $\triangle POQ$ has area $21$ square units.

(i) Express $y$ in terms of $x$

Dividing $9x^2+16y^2=144$ by $144$ gives $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ , so $a=4,\ b=3$ . Then $y^2=\dfrac{9(16-x^2)}{16}$ and, taking the upper half,

y=\frac34\sqrt{16-x^2}.

(ii) Integrate

Using $\displaystyle\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa+C$ with $a=4$ :

\int\frac34\sqrt{16-x^2}\,dx=\frac34\left(\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\frac x4\right)+C=\frac38\left(x\sqrt{16-x^2}+16\sin^{-1}\frac x4\right)+C.

Part (a) — Area of the elliptical ground

The racing track is built around the ground, so the ground itself is exactly the region enclosed by the given ellipse. By symmetry the total area is four times the first-quadrant area:

$A=4\int_0^4\frac34\sqrt{16-x^2}\,dx=3\left[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\frac x4\right]_0^4.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set 65/3/11 markMCQ
$\int_{1}^{3} \sqrt{9-y^2}\, dy$
›Reveal solutionSolution
The problem asks for the integral representing the area of a shaded region of a circle. Assuming the shaded region is the quarter circle in the first quadrant of $x^2+y^2=9$ , its area is given by $\boxed{\int_{0}^{3} \sqrt{9-x^2}\, dx}$ .

The core concept here is using definite integrals to calculate the area under a curve. When we have a region bounded by a curve $y=f(x)$ , the x-axis, and vertical lines $x=a$ and $x=b$ , the area is given by $\int_a^b f(x)\, dx$ . Similarly, if the region is bounded by a curve $x=g(y)$ , the y-axis, and horizontal lines $y=c$ and $y=d$ , the area is $\int_c^d g(y)\, dy$ .

The expressions in the options, $\sqrt{9-x^2}$ and $\sqrt{9-y^2}$ , immediately point to the equation of a circle. The general equation of a circle centered at the origin with radius $r$ is $x^2+y^2=r^2$ . Comparing this with $9-x^2$ or $9-y^2$ , we see that $r^2=9$ , which means the radius $r=3$ .

For the upper half of this circle, we can express $y$ as a function of $x$ : $y^2 = 9-x^2 \implies y = \sqrt{9-x^2}$ (taking the positive root for the upper half).

For the right half of this circle, we can express $x$ as a function of $y$ : $x^2 = 9-y^2 \implies x = \sqrt{9-y^2}$ (taking the positive root for the right half).

Since the figure is not provided, we must infer the shaded region from the given options. Options (C) and (D) involve integration from $0$ to $3$ , which is the radius of the circle. This strongly suggests that the shaded region is either a quarter circle or a semi-circle.

Identify the circle's equation and radius:

The terms $\sqrt{9-x^2}$ and $\sqrt{9-y^2}$ indicate that the circle has the equation $x^2+y^2=9$ . This is a circle centered at the origin $(0,0)$ with a radius $r=3$ .

Interpret the shaded region based on options:

Option (C) is $\int_{0}^{3} \sqrt{9-x^2}\, dx$ . This integral represents the area under the curve $y=\sqrt{9-x^2}$ (the upper semi-circle) from $x=0$ to $x=3$ . This region is precisely the quarter circle located in the first quadrant.

Option (D) is $2 \int_{0}^{3} \sqrt{9-x^2}\, dx$ . This would be twice the area of the quarter circle, meaning it represents the area of the entire upper semi-circle (from $x=-3$ to $x=3$ , or by symmetry, $2 \times$ area from $x=0$ to $x=3$ ).

Option (A) is $\int_{1}^{3} \sqrt{9-y^2}\, dy$ . This represents the area under the curve $x=\sqrt{9-y^2}$ (the right semi-circle) from $y=1$ to $y=3$ . This is a specific segment of the quarter circle, not the entire quarter circle.

Option (B) is $2 \int_{1}^{3} \sqrt{9-y^2}\, dy$ . This would be twice the area in (A), representing a horizontal strip of the circle symmetric about the y-axis. …
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CBSE 2026Set ANNUAL1 markMCQ
$y=\sin x$
(a) 3
(b) 4
(c) 2
(d) 5
›Reveal solutionSolution

The area under one arch of $y=\sin x$ from $x=0$ to $x=\pi$ is $2$ square units.

Since $\sin x\ge0$ throughout $[0,\pi]$ , the area is simply the definite integral:

$A=\int_0^{\pi}\sin x\,dx=\big[-\cos x\big]_0^{\pi}$

…

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CBSE 2026Set ANNUAL1 mark
$x^2 + y^2 = a^2$
›Reveal solutionSolution

By symmetry, the area of the full circle is 4 times the area in the first quadrant, which is found by integration.

…

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CBSE 2025Set 65/4/11 markMCQ
$y = \sqrt{x}$
›Reveal solutionSolution
The region is the area under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ , which is a standard definite integral. The area equals $\frac{16}{3}$ square units, so the correct option is (C).

The problem asks for the area enclosed by the curve $y = \sqrt{x}$ , the vertical lines $x = 0$ and $x = 4$ , and the x-axis. This is a classic "area under a curve" problem — the region is bounded above by the curve, below by the x-axis, and on the sides by two vertical lines. The key idea is that the area between a curve $y = f(x)$ and the x-axis from $x = a$ to $x = b$ is given by the definite integral $\int_a^b f(x) \, dx$ , provided $f(x) \ge 0$ on that interval. Here, $\sqrt{x}$ is non-negative for $x \ge 0$ , so we can directly integrate.

Watch out

A common mistake is to confuse the area under $y = \sqrt{x}$ with the area under $y = x^2$ or to misapply the power rule. Always check the exponent: $\sqrt{x} = x^{1/2}$ , not $x^2$ .

Let’s work through the calculation step by step.

Set up the integral. The region is bounded by $x = 0$ on the left and $x = 4$ on the right. The curve is $y = \sqrt{x}$ , and the lower boundary is the x-axis ( $y = 0$ ). So the area $A$ is:

$A = \int_{0}^{4} \sqrt{x} \, dx$

Rewrite the integrand. Recall that $\sqrt{x} = x^{1/2}$ . This makes the power rule for integration straightforward:

$A = \int_{0}^{4} x^{1/2} \, dx$

Apply the power rule. …
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CBSE 2025Set ANNUAL1 markMCQ
$y=e^{x}$
(i) $e(e^{2}-1)$
(ii) $e^{3}-1$
(iii) $e(1-e^{2})$
(iv) $e^{2}(1-e)$
›Reveal solutionSolution

Area under $y=e^x$ from $x=1$ to $x=3$ is $\displaystyle\int_1^3 e^x\,dx$ .

Since $y=e^x > 0$ throughout $[1,3]$ , the region between the curve and the $X$ -axis has area: …

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CBSE 2025Set ANNUAL1 markMCQ
$x$
(a) $1$
(b) $0$
(c) $-1$
(d) $2$
›Reveal solutionSolution

The required area is $\int_0^{\pi/2}\cos x\,dx$ .

…

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CBSE 2025Set ANNUAL1 mark
$x^2+y^2=4$
›Reveal solutionSolution

The full circle $x^2+y^2=4$ has radius $2$ ; the first-quadrant portion is one quarter of the full circle.

Full circle area $=\pi r^2=\pi(2)^2=4\pi$ .

Area in the first quadrant $=\dfrac{1}{4}\times4\pi=\pi$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$x^2 + y^2 = 9$
(a) $9\pi$
(b) $\dfrac{3\pi}{4}$
(c) $\dfrac{9\pi}{4}$
(d) $3\pi$
›Reveal solutionSolution

The circle $x^2+y^2=9$ has radius 3; the first-quadrant region is one quarter of the full circle.

Full circle area $= \pi r^2 = \pi(3)^2 = 9\pi$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$x^2 + y^2 = 2$
(a) $4\pi$ sq. units
(b) $2\sqrt{2}\pi$ sq. units
(c) $4\pi^2$ sq. units
(d) $2\pi$ sq. units
›Reveal solutionSolution

$2\pi$ sq. units — option (d).

The circle $x^2+y^2=2$ has radius $R=\sqrt2$ (comparing with $x^2+y^2=R^2$ ).

…

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CBSE 2022Set ANNUAL1 markMCQ
$y=x$
(a) 8 sq. units
(b) 4 sq. units
(c) 2 sq. units
(d) 1 sq. unit
›Reveal solutionSolution

The area under a straight line $y=x$ between two vertical lines is a definite integral, here it also equals the area of a trapezium.

…

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CBSE 2022Set ANNUAL1 mark
$Q. The area bounded by the curve y = 2x between x = 0, x = 2 and x-axis is ...$
›Reveal solutionSolution

Integrate y = 2x from x = 0 to x = 2 to get the area under the line.

The curve $y=2x$ is a straight line through the origin. The area bounded by it, the x-axis, and the ordinates $x=0$ and $x=2$ is

$\text{Area} = \displaystyle\int_0^2 2x\,dx = \Big[x^2\Big]_0^2 = 4-0 = 4$ square units.

…

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CBSE 2020Set HE8231 mark
$x^2+y^2=4$
›Reveal solutionSolution

The required area is $\pi$ square units — a quarter of the circle $x^2+y^2=4$ .

The circle $x^2+y^2=4$ has centre $(0,0)$ and radius $2$ . In the first quadrant, $y=\sqrt{4-x^2}$ for $0\le x\le2$ . The line $y=2$ touches this circle at the single point $(0,2)$ (it is tangent there, since the circle's topmost point is $(0,2)$ ), so bounding the region additionally by $y=2$ and $x=0$ does not remove or add any area beyond the quarter-disc that already lies between the curve and the two axes.

…

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Q.A racing track is build around an elliptical ground whose equation is given by $9x^2 + 16y^2 = 144$ . The width of the track is 3 m as shown below. Based on given information, answer the following questions:

Part (a)Concept understanding — Area of Ellipse

Area of an Ellipse

Finding it by integration

The intuition

Part (b)Concept understanding — Area Under Curve

Area Under a Curve

The core idea: area as a limit of strips

How to compute it

Cases you must handle carefully

Area with respect to the y-axis

Part (a) — Area of the elliptical ground (excluding the track)

(i) Express $y$ in terms of $x$

(ii) Integrate

Part (a) — Area of the elliptical ground

Q.A racing track is build around an elliptical ground whose equation is given by 9x2+16y2=1449x^2 + 16y^2 = 1449x2+16y2=144. The width of the track is 3 m as shown below. Based on given information, answer the following questions:

(i) Express yyy in terms of xxx

(ii) Integrate

Part (a) — Area of the elliptical ground

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Q.A racing track is build around an elliptical ground whose equation is given by $9x^2 + 16y^2 = 144$ . The width of the track is 3 m as shown below. Based on given information, answer the following questions:

(i) Express $y$ in terms of $x$

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