Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 10 – Conic Sections (Exercise 10.3: Ellipse) following the latest NCERT and CBSE syllabus (2025–2026). This exercise explains the definition and standard equation of an ellipse, along with important terms such as foci, major axis, minor axis, eccentricity, and center. Students will also learn how to derive and solve problems on equations of ellipses in standard and general form. Each solution is presented step-by-step to help strengthen concepts in coordinate geometry and prepare for CBSE board exams, JEE Main, JEE Advanced, NDA, and CUET. Click the print button to download study material and notes.
NCERT Question.10 : Find the Equation of Ellipse with
Vertices $(\pm 5, 0)$, Foci $(\pm 4, 0)$.
Solution:
Since the vertices are on the x-axis, the equation will be of the form
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where $a$ is the semi-major axis $(a^2 > b^2)$.
Given: $a = 5$, $c = 4$
From the relation,
$$c^2 = a^2 – b^2$$
$$b^2 = a^2 – c^2 = 25 – 16 = 9$$
Hence,
$$a^2 = 25, \quad b^2 = 9$$
The required equation of the ellipse is
$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$
NCERT Question.11 : Find the Equation of Ellipse with
Vertices $(0, \pm 13)$, Foci $(0, \pm 5)$
Solution:
Since the vertices are on the y-axis, the equation will be of the form
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where $b$ is the semi-major axis $(a^2 < b^2)$.
Given: $b = 13$, $c = 5$
From the relation,
$$c^2 = b^2 – a^2$$
$$a^2 = b^2 – c^2 = 169 – 25 = 144$$
Hence,
$$a^2 = 144, \quad b^2 = 169$$
The required equation of the ellipse is
$$\frac{x^2}{144} + \frac{y^2}{169} = 1$$
NCERT Question.12 : Find the Equation of Ellipse with
Vertices $(\pm 6, 0)$, Foci $(\pm 4, 0)$
Solution:
Since the vertices are on the x-axis, the equation is
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where $a^2 > b^2$.
Given: $a = 6$, $c = 4$
Then,
$$b^2 = a^2 – c^2 = 36 – 16 = 20$$
Hence,
$$a^2 = 36, \quad b^2 = 20$$
The required equation is
$$\frac{x^2}{36} + \frac{y^2}{20} = 1$$
NCERT Question.13 : Find the Equation of Ellipse with ends of major axis $(\pm 3, 0)$ and ends of minor axis $(0, \pm 2)$.
Solution:
The equation of the ellipse is
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where $a = 3$, $b = 2$.
Hence,
$$a^2 = 9, \quad b^2 = 4$$
The required equation is
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
NCERT Question.14 : Find the Equation of Ellipse with ends of major axis $(0, \pm \sqrt{5})$ and ends of minor axis $(\pm 1, 0)$.
Solution:
Since the major axis lies on the y-axis,
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where $b^2 > a^2$.
Given: $a = 1$, $b = \sqrt{5}$
Hence,
$$a^2 = 1, \quad b^2 = 5$$
The required equation is
$$\frac{x^2}{1} + \frac{y^2}{5} = 1$$
NCERT Question.15 : Find the Equation of Ellipse with Length of major axis $= 26$ and Foci $(\pm 5, 0)$.
Solution:
Since the foci are on the x-axis,
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Given: $2a = 26 \Rightarrow a = 13$, $c = 5$
Then,
$$b^2 = a^2 – c^2 = 169 – 25 = 144$$
Hence,
$$a^2 = 169, \quad b^2 = 144$$
The required equation is
$$\frac{x^2}{169} + \frac{y^2}{144} = 1$$
NCERT Question.16 : Find the Equation of Ellipse with Length of minor axis $= 16$ and Foci $(0, \pm 6)$.
Solution:
Since the foci are on the y-axis,
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where $a^2 < b^2$.
Given: $2a = 16 \Rightarrow a = 8$, $c = 6$
Then,
$$b^2 = c^2 + a^2 = 36 + 64 = 100$$
Hence,
$$a^2 = 64, \quad b^2 = 100$$
The required equation is
$$\frac{x^2}{64} + \frac{y^2}{100} = 1$$
NCERT Question.17 : Find the Equation of Ellipse with Foci $(\pm 3, 0)$ and $a = 4$.
Solution:
Since the foci are on the x-axis,
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Given: $a = 4$, $c = 3$
Then,
$$b^2 = a^2 – c^2 = 16 – 9 = 7$$
Hence,
$$a^2 = 16, \quad b^2 = 7$$
The required equation is
$$\frac{x^2}{16} + \frac{y^2}{7} = 1$$
NCERT Question.18 : Find the Equation of Ellipse
with $b = 3$, $c = 4$, centre at the origin and foci on x-axis.
Solution:
Equation of ellipse:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
From $c^2 = a^2 – b^2$,
$$a^2 = b^2 + c^2 = 9 + 16 = 25$$
Hence,
$$a^2 = 25, \quad b^2 = 9$$
Required equation:
$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$
NCERT Question.19 : Find the Equation of Ellipse with
Centre $(0, 0)$, major axis on y-axis, passes through $(3, 2)$ and $(1, 6)$.
Solution:
Given Equation:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Substituting $(3, 2)$:
$$\frac{9}{a^2} + \frac{4}{b^2} = 1 \quad (1)$$
Substituting $(1, 6)$:
$$\frac{1}{a^2} + \frac{36}{b^2} = 1 \quad (2)$$
Multiplying (1) by 9 and subtracting (2):
$$\frac{80}{a^2} = 8 \Rightarrow a^2 = 10$$
Substitute in (1):
$$\frac{9}{10} + \frac{4}{b^2} = 1 \Rightarrow \frac{4}{b^2} = \frac{1}{10} \Rightarrow b^2 = 40$$
Hence,
$$a^2 = 10, \quad b^2 = 40$$
Equation:
$$\frac{x^2}{10} + \frac{y^2}{40} = 1$$
NCERT Question.20 : Find the Equation of Ellipse with Major axis on x-axis, passes through $(4, 3)$ and $(6, 2)$
Solution:
Given Equation :
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Substituting $(4, 3)$:
$$\frac{16}{a^2} + \frac{9}{b^2} = 1 \quad (1)$$
Substituting $(6, 2)$:
$$\frac{36}{a^2} + \frac{4}{b^2} = 1 \quad (2)$$
Multiplying (2) by 9 and subtracting (1) × 4:
$$\frac{260}{a^2} = 5 \Rightarrow a^2 = 52$$
Substitute in (1):
$$\frac{16}{52} + \frac{9}{b^2} = 1 \Rightarrow \frac{9}{b^2} = \frac{36}{52} \Rightarrow b^2 = 13$$
Hence,
$$a^2 = 52, \quad b^2 = 13$$
The required equation of the ellipse is
$$\frac{x^2}{52} + \frac{y^2}{13} = 1$$
Explore more: Learn ellipse derivations and equations step-by-step with Anand Classes — perfect for Class 11 Maths, JEE, and CUET geometry preparation.
