NCERT Class 11 Mathematics — Conic Sections, Exercise 10.3. All 20 questions solved.

An ellipse is the set of points whose distances from two fixed points (the foci) have a constant sum. With centre at the origin there are two standard forms, and $a$ always denotes the semi-major axis, so that $a > b$:

Form	Major axis	Foci	Vertices
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$	along $x$	$(\pm c, 0)$	$(\pm a, 0)$
$\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$	along $y$	$(0, \pm c)$	$(0, \pm a)$

with

$$c^2 = a^2 – b^2, \qquad e = \frac ca, \qquad \text{latus rectum} = \frac{2b^2}{a}$$

Lengths: major axis $= 2a$, minor axis $= 2b$. Note $0 < e < 1$ for every ellipse.

Key insight. The larger denominator is $a^2$, and it names the major axis — not the order in which $x$ and $y$ happen to be written. In question 2, $\tfrac{x^2}{4} + \tfrac{y^2}{25} = 1$ has $25$ under $y^2$, so the major axis is the $y$-axis and the foci are on it. Reading this off correctly is the first step of every one of questions 1 to 9, and getting it wrong swaps every subsequent answer.

In questions 1 to 9, find the coordinates of the foci, the vertices, the lengths of the major and minor axes, the eccentricity and the length of the latus rectum of the ellipse.

Question 1

$\dfrac{x^2}{36} + \dfrac{y^2}{16} = 1$

Solution. The larger denominator, $36$, sits under $x^2$, so the major axis is along the $x$-axis:

$$a^2 = 36 \Rightarrow a = 6, \qquad b^2 = 16 \Rightarrow b = 4$$

$$c^2 = a^2 – b^2 = 36 – 16 = 20 \quad\Longrightarrow\quad c = 2\sqrt5$$

$$e = \frac ca = \frac{2\sqrt5}{6} = \frac{\sqrt5}{3}, \qquad \text{latus rectum} = \frac{2b^2}{a} = \frac{32}{6} = \frac{16}{3}$$

Question 2

$\dfrac{x^2}{4} + \dfrac{y^2}{25} = 1$

Solution. Here $25 > 4$, so $a^2 = 25$ and the major axis is along the $y$-axis:

$$a = 5, \qquad b = 2, \qquad c = \sqrt{25 – 4} = \sqrt{21}$$

$$e = \frac{\sqrt{21}}{5}, \qquad \text{latus rectum} = \frac{2(4)}{5} = \frac85$$

Question 3

$\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$

Solution. $a = 4$, $b = 3$, $c = \sqrt{16-9} = \sqrt7$; major axis along $x$.

$$e = \frac{\sqrt7}{4}, \qquad \text{latus rectum} = \frac{2(9)}{4} = \frac92$$

Question 4

$\dfrac{x^2}{25} + \dfrac{y^2}{100} = 1$

Solution. $a^2 = 100$ under $y^2$, so the major axis is along $y$: $a = 10$, $b = 5$, $c = \sqrt{100-25} = \sqrt{75} = 5\sqrt3$.

$$e = \frac{5\sqrt3}{10} = \frac{\sqrt3}{2}, \qquad \text{latus rectum} = \frac{2(25)}{10} = 5$$

Question 5

$\dfrac{x^2}{49} + \dfrac{y^2}{36} = 1$

Solution. $a = 7$, $b = 6$, $c = \sqrt{49-36} = \sqrt{13}$; major axis along $x$.

$$e = \frac{\sqrt{13}}{7}, \qquad \text{latus rectum} = \frac{2(36)}{7} = \frac{72}{7}$$

Question 6

$\dfrac{x^2}{100} + \dfrac{y^2}{400} = 1$

Solution. $a^2 = 400$, so $a = 20$, $b = 10$, $c = \sqrt{400-100} = \sqrt{300} = 10\sqrt3$; major axis along $y$.

$$e = \frac{10\sqrt3}{20} = \frac{\sqrt3}{2}, \qquad \text{latus rectum} = \frac{2(100)}{20} = 10$$

Question 7

$36x^2 + 4y^2 = 144$

Solution. Divide throughout by $144$ to reach standard form — this step is essential, since the standard formulae assume the right-hand side is $1$:

$$\frac{36x^2}{144} + \frac{4y^2}{144} = 1 \quad\Longrightarrow\quad \frac{x^2}{4} + \frac{y^2}{36} = 1$$

Now $a^2 = 36$, $b^2 = 4$, so $a = 6$, $b = 2$, $c = \sqrt{36-4} = \sqrt{32} = 4\sqrt2$; major axis along $y$.

$$e = \frac{4\sqrt2}{6} = \frac{2\sqrt2}{3}, \qquad \text{latus rectum} = \frac{2(4)}{6} = \frac43$$

Question 8

$16x^2 + y^2 = 16$

Solution. Dividing by $16$:

$$\frac{x^2}{1} + \frac{y^2}{16} = 1$$

$a = 4$, $b = 1$, $c = \sqrt{16-1} = \sqrt{15}$; major axis along $y$.

$$e = \frac{\sqrt{15}}{4}, \qquad \text{latus rectum} = \frac{2(1)}{4} = \frac12$$

Question 9

$4x^2 + 9y^2 = 36$

Solution. Dividing by $36$:

$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

$a = 3$, $b = 2$, $c = \sqrt{9-4} = \sqrt5$; major axis along $x$.

$$e = \frac{\sqrt5}{3}, \qquad \text{latus rectum} = \frac{2(4)}{3} = \frac83$$

In questions 10 to 20, find the equation of the ellipse that satisfies the given conditions.

Question 10

Vertices $(\pm5, 0)$, foci $(\pm4, 0)$.

Solution. Both lie on the $x$-axis, so the major axis is along $x$ with $a = 5$ and $c = 4$:

$$b^2 = a^2 – c^2 = 25 – 16 = 9$$

$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$

Question 11

Vertices $(0, \pm13)$, foci $(0, \pm5)$.

Solution. Major axis along $y$: $a = 13$, $c = 5$, so $b^2 = 169 – 25 = 144$.

$$\frac{x^2}{144} + \frac{y^2}{169} = 1$$

Note that $a^2$ goes under $y^2$ — the larger denominator belongs to the major axis.

Question 12

Vertices $(\pm6, 0)$, foci $(\pm4, 0)$.

Solution. $a = 6$, $c = 4$, so $b^2 = 36 – 16 = 20$.

$$\frac{x^2}{36} + \frac{y^2}{20} = 1$$

Question 13

Ends of major axis $(\pm3, 0)$, ends of minor axis $(0, \pm2)$.

Solution. Read $a$ and $b$ directly: $a = 3$ (on the $x$-axis) and $b = 2$.

$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

Question 14

Ends of major axis $\left(0, \pm\sqrt5\right)$, ends of minor axis $(\pm1, 0)$.

Solution. The major axis is along $y$ with $a = \sqrt5$, so $a^2 = 5$; and $b = 1$.

$$\frac{x^2}{1} + \frac{y^2}{5} = 1$$

Question 15

Length of major axis $26$, foci $(\pm5, 0)$.

Solution. Major axis $= 2a = 26$, so $a = 13$. The foci on the $x$-axis give $c = 5$:

$$b^2 = 169 – 25 = 144$$

$$\frac{x^2}{169} + \frac{y^2}{144} = 1$$

Question 16

Length of minor axis $16$, foci $(0, \pm6)$.

Solution. Minor axis $= 2b = 16$, so $b = 8$. The foci on the $y$-axis give $c = 6$ and place the major axis along $y$:

$$a^2 = b^2 + c^2 = 64 + 36 = 100$$

Note the rearrangement — here $a$ is the unknown, so $c^2 = a^2 – b^2$ is used the other way round.

$$\frac{x^2}{64} + \frac{y^2}{100} = 1$$

Question 17

Foci $(\pm3, 0)$, $a = 4$.

Solution. $c = 3$ and $a = 4$ with the major axis along $x$:

$$b^2 = 16 – 9 = 7$$

$$\frac{x^2}{16} + \frac{y^2}{7} = 1$$

Question 18

$b = 3$, $c = 4$, centre at the origin, foci on the $x$-axis.

Solution.

$$a^2 = b^2 + c^2 = 9 + 16 = 25$$

Foci on the $x$-axis means $a^2$ goes under $x^2$:

$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$

Question 19

Centre at $(0,0)$, major axis on the $y$-axis, and passing through the points $(3, 2)$ and $(1, 6)$.

Solution. With the major axis on the $y$-axis the equation is

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

Substituting both points gives two equations. Writing $u = \dfrac{1}{b^2}$ and $v = \dfrac{1}{a^2}$ turns them into a linear system, which is the trick that makes this manageable:

$$9u + 4v = 1 \qquad (1)$$ $$u + 36v = 1 \qquad (2)$$

From (2), $u = 1 – 36v$. Substituting into (1):

$$9(1 – 36v) + 4v = 1 \quad\Longrightarrow\quad 9 – 324v + 4v = 1 \quad\Longrightarrow\quad -320v = -8$$

$$v = \frac{1}{40} \quad\Longrightarrow\quad a^2 = 40$$

$$u = 1 – \frac{36}{40} = \frac{1}{10} \quad\Longrightarrow\quad b^2 = 10$$

Check that $a^2 > b^2$: $40 > 10$ ✓, consistent with the major axis being on $y$.

$$\frac{x^2}{10} + \frac{y^2}{40} = 1$$

Question 20

Major axis on the $x$-axis and passing through the points $(4, 3)$ and $(6, 2)$.

Solution. The form is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. With $u = \dfrac{1}{a^2}$ and $v = \dfrac{1}{b^2}$:

$$16u + 9v = 1 \qquad (1)$$ $$36u + 4v = 1 \qquad (2)$$

Multiply (1) by $4$ and (2) by $9$ to eliminate $v$:

$$64u + 36v = 4, \qquad 324u + 36v = 9$$

Subtracting: $260u = 5$, so $u = \dfrac{1}{52}$ and $a^2 = 52$.

From (1): $9v = 1 – \dfrac{16}{52} = \dfrac{36}{52}$, so $v = \dfrac{1}{13}$ and $b^2 = 13$.

$$\frac{x^2}{52} + \frac{y^2}{13} = 1 \quad\text{or, cleared of fractions,}\quad x^2 + 4y^2 = 52$$

Common mistakes

• Assuming $a^2$ always sits under $x^2$. It sits under whichever variable corresponds to the major axis. In questions 2, 4, 6, 7 and 8 that is $y$.
• Questions 7 to 9, forgetting to divide by the constant. $36x^2 + 4y^2 = 144$ is not in standard form until the right side is $1$. Reading $a^2 = 36$ off the undivided equation is the commonest error in this exercise.
• Using $c^2 = a^2 + b^2$. That is the hyperbola relation. For an ellipse $c^2 = a^2 – b^2$, which is why $c < a$ and $e < 1$.
• Question 16, computing $a^2 = b^2 – c^2$. Here $a$ is unknown, so the relation must be rearranged as $a^2 = b^2 + c^2$. A negative result is the signal that the rearrangement went the wrong way.
• Giving the major axis length as $a$ rather than $2a$. The axes are $2a$ and $2b$; the semi-axes are $a$ and $b$.
• Questions 19 and 20, solving the simultaneous equations in $a^2$ and $b^2$ directly. Substituting $u = \tfrac{1}{a^2}$ and $v = \tfrac{1}{b^2}$ makes the system linear and takes three lines instead of a page.
• Question 19, not checking $a^2 > b^2$. If the numbers come out the other way round, the major axis is not where you assumed and the form must be swapped.

Practise next

• Exercise 10.4 — hyperbolas, whose formulae are the ellipse’s with the sign of $b^2$ reversed: $c^2 = a^2 + b^2$ and $e > 1$.
• Miscellaneous Exercise on Chapter 10 — applied problems on parabolic reflectors, arches and elliptical orbits.