Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 10 – Conic Sections (Exercise 10.4: Hyperbola) strictly based on the latest NCERT and CBSE syllabus (2025–2026). This exercise focuses on the definition, standard equation, and properties of a hyperbola, including its transverse axis, conjugate axis, foci, vertices, and eccentricity. Students will learn how to derive the equation of a hyperbola and solve numerical problems related to its geometrical properties. Each question is explained step-by-step to make the learning process easy and effective for CBSE board exams, JEE Main, JEE Advanced, NDA, and CUET. Click the print button to download study material and notes.
NCERT Question.1 : Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
$$\frac{x^2}{16} – \frac{y^2}{9} = 1$$
Solution:
Comparing the given equation with
$$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$
We conclude that the transverse axis is along the x-axis.
Hence,
$$a^2 = 16, \quad b^2 = 9$$
$$a = \pm 4, \quad b = \pm 3$$
Foci:
Foci are $(c, 0)$ and $(-c, 0)$
$$c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$
So, the foci are $(5, 0)$ and $(-5, 0)$
Vertices:
Vertices are $(a, 0)$ and $(-a, 0)$
Hence, $(4, 0)$ and $(-4, 0)$
Eccentricity:
$$e = \frac{c}{a} = \frac{5}{4}$$
Length of Latus Rectum:
$$= \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2}$$
NCERT Question.2 : Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
$$\frac{y^2}{9} – \frac{x^2}{27} = 1$$
Solution:
Comparing the given equation with
$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$
We conclude that the transverse axis is along the y-axis.
Hence,
$$a^2 = 9, \quad b^2 = 27$$
$$a = \pm 3, \quad b = \pm 3\sqrt{3}$$
Foci:
Foci are $(0, c)$ and $(0, -c)$
$$c = \sqrt{a^2 + b^2} = \sqrt{9 + 27} = \sqrt{36} = 6$$
So, the foci are $(0, 6)$ and $(0, -6)$
Vertices:
Vertices are $(0, a)$ and $(0, -a)$
Hence, $(0, 3)$ and $(0, -3)$
Eccentricity:
$$e = \frac{c}{a} = \frac{6}{3} = 2$$
Length of Latus Rectum:
$$= \frac{2b^2}{a} = \frac{2 \times 27}{3} = 18$$
NCERT Question.3 : Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
$$9y^2 – 4x^2 = 36$$
Solution:
Dividing both sides by $36$, we get
$$\frac{y^2}{4} – \frac{x^2}{9} = 1$$
Comparing the given equation with
$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$
We conclude that the transverse axis is along the y-axis.
Hence,
$$a^2 = 4, \quad b^2 = 9$$
$$a = \pm 2, \quad b = \pm 3$$
Foci:
Foci are $(0, c)$ and $(0, -c)$
$$c = \sqrt{a^2 + b^2} = \sqrt{4 + 9} = \sqrt{13}$$
So, the foci are $(0, \sqrt{13})$ and $(0, -\sqrt{13})$
Vertices:
Vertices are $(0, a)$ and $(0, -a)$
Hence, $(0, 2)$ and $(0, -2)$
Eccentricity:
$$e = \frac{c}{a} = \frac{\sqrt{13}}{2}$$
Length of Latus Rectum:
$$= \frac{2b^2}{a} = \frac{2 \times 9}{2} = 9$$
NCERT Question 4 : Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
$$16x^2 – 9y^2 = 576$$
Solution:
Dividing both sides by $576$, we get
$$\frac{x^2}{36} – \frac{y^2}{64} = 1$$
Comparing the given equation with
$$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$
We conclude that the transverse axis is along the x-axis.
Hence,
$$a^2 = 36, \quad b^2 = 64$$
$$a = \pm 6, \quad b = \pm 8$$
Foci:
Foci are $(c, 0)$ and $(-c, 0)$
$$c = \sqrt{a^2 + b^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
So, the foci are $(10, 0)$ and $(-10, 0)$
Vertices:
Vertices are $(a, 0)$ and $(-a, 0)$
Hence, $(6, 0)$ and $(-6, 0)$
Eccentricity:
$$e = \frac{c}{a} = \frac{10}{6} = \frac{5}{3}$$
Length of Latus Rectum:
$$= \frac{2b^2}{a} = \frac{2 \times 64}{6} = \frac{64}{3}$$
NCERT Question 5 : Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
$$5y^2 – 9x^2 = 36$$
Solution:
Dividing both sides by $36$, we get
$$\dfrac{y^2}{\dfrac{36}{5}} – \dfrac{x^2}{4} = 1$$
Comparing with
$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$
We conclude that the transverse axis is along the y-axis.
Hence,
$$a^2 = \frac{36}{5}, \quad b^2 = 4$$
$$a = \pm \frac{6}{\sqrt{5}}, \quad b = \pm 2$$
Foci:
Foci are $(0, c)$ and $(0, -c)$
$$c = \sqrt{a^2 + b^2} = \sqrt{\frac{36}{5} + 4} = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}}$$
So, the foci are $(0, \frac{2\sqrt{14}}{\sqrt{5}})$ and $(0, -\frac{2\sqrt{14}}{\sqrt{5}})$
Vertices:
Vertices are $(0, a)$ and $(0, -a)$
Hence, $(0, \frac{6}{\sqrt{5}})$ and $(0, -\frac{6}{\sqrt{5}})$
Eccentricity:
$$e = \frac{c}{a} = \frac{\frac{2\sqrt{14}}{\sqrt{5}}}{\frac{6}{\sqrt{5}}} = \frac{\sqrt{14}}{3}$$
Length of Latus Rectum:
$$= \frac{2b^2}{a} = \frac{2 \times 4}{\frac{6}{\sqrt{5}}} = \frac{4\sqrt{5}}{3}$$
NCERT Question 6 : Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
$$49y^2 – 16x^2 = 784$$
Solution:
Dividing both sides by $784$, we get
$$\frac{y^2}{16} – \frac{x^2}{49} = 1$$
Comparing the given equation with
$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$
We conclude that the transverse axis is along the y-axis.
Hence,
$$a^2 = 16, \quad b^2 = 49$$
$$a = \pm 4, \quad b = \pm 7$$
Foci:
Foci are $(0, c)$ and $(0, -c)$
$$c = \sqrt{a^2 + b^2} = \sqrt{16 + 49} = \sqrt{65}$$
So, the foci are $(0, \sqrt{65})$ and $(0, -\sqrt{65})$
Vertices:
Vertices are $(0, a)$ and $(0, -a)$
Hence, $(0, 4)$ and $(0, -4)$
Eccentricity:
$$e = \frac{c}{a} = \frac{\sqrt{65}}{4}$$
Length of Latus Rectum:
$$= \frac{2b^2}{a} = \frac{2 \times 49}{4} = \frac{49}{2}$$
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