Conic Sections (Hyperbola) NCERT Solutions Exercise 10.4 Class 11 Math (Set-2)

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 7 : Find the equation of the hyperbola with vertices $(\pm 2, 0)$ and foci $(\pm 3, 0)$.

Solution:
Since the foci lie on the x-axis, the equation of the hyperbola is of the form
$$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$

Given, vertices $(\pm 2, 0)$ and foci $(\pm 3, 0)$

So,
$$a = 2, \quad c = 3$$

We know that
$$c^2 = a^2 + b^2$$

$$b^2 = c^2 – a^2 = 9 – 4 = 5$$

Hence, the equation is
$$\frac{x^2}{4} – \frac{y^2}{5} = 1$$

NCERT Question 8 : Find the equation of the hyperbola with vertices $(0, \pm 5)$ and foci $(0, \pm 8)$.

Solution:
Since the foci lie on the y-axis, the equation of the hyperbola is of the form
$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$

Given, vertices $(0, \pm 5)$ and foci $(0, \pm 8)$

So,
$$a = 5, \quad c = 8$$

We know that
$$b^2 = c^2 – a^2 = 64 – 25 = 39$$

Hence, the equation is
$$\frac{y^2}{25} – \frac{x^2}{39} = 1$$

NCERT Question 9 : Find the equation of the hyperbola with vertices $(0, \pm 3)$ and foci $(0, \pm 5)$.

Solution:
Since the foci lie on the y-axis, the equation of the hyperbola is of the form
$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$

Given, vertices $(0, \pm 3)$ and foci $(0, \pm 5)$

So,
$$a = 3, \quad c = 5$$

We know that
$$b^2 = c^2 – a^2 = 25 – 9 = 16$$

Hence, the equation is
$$\frac{y^2}{9} – \frac{x^2}{16} = 1$$

NCERT Question 10 : Find the equation of the hyperbola whose foci are $(\pm 5, 0)$ and whose transverse axis is of length $8$.

Solution:
Since the foci lie on the x-axis, the equation of the hyperbola is of the form
$$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$

Given, $c = 5$ and length of the transverse axis $= 8 \Rightarrow 2a = 8$
So,
$$a = 4$$

Now,
$$b^2 = c^2 – a^2 = 25 – 16 = 9$$

Hence, the equation is
$$\frac{x^2}{16} – \frac{y^2}{9} = 1$$

NCERT Question 11 : Find the equation of the hyperbola whose foci are $(0, \pm 13)$ and the conjugate axis is of length $24$.

Solution:
Since the foci lie on the y-axis, the equation of the hyperbola is of the form
$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$

Given, $c = 13$ and length of conjugate axis $= 24 \Rightarrow 2b = 24$
So,
$$b = 12$$

Now,
$$a^2 = c^2 – b^2 = 169 – 144 = 25$$

Hence, the equation is
$$\frac{y^2}{25} – \frac{x^2}{144} = 1$$

NCERT Question 12 : Find the equation of the hyperbola whose foci are $(\pm 3\sqrt{5}, 0)$ and whose latus rectum is of length $8$.

Solution:
Since the foci lie on the x-axis, the equation of the hyperbola is of the form
$$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$

Given, $c = 3\sqrt{5}$ and latus rectum $= 8$.

We know,
$$\text{Latus rectum} = \frac{2b^2}{a} = 8$$
So,
$$b^2 = 4a \quad \text{…(1)}$$

Also,
$$c^2 = a^2 + b^2$$

$$45 = a^2 + 4a$$

$$a^2 + 4a – 45 = 0$$

Solving,
$$(a + 9)(a – 5) = 0$$

Since $a > 0$, we take $a = 5$.
Substitute in (1):
$$b^2 = 4a = 20$$

Hence,
$$a^2 = 25, \quad b^2 = 20$$

The equation is
$$\frac{x^2}{25} – \frac{y^2}{20} = 1$$

NCERT Question 13 : Find the equation of the hyperbola whose foci are $(\pm 4, 0)$ and whose latus rectum is of length $12$.

Solution:
Since the foci lie on the x-axis, the equation of the hyperbola is of the form
$$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$

Given, $c = 4$ and latus rectum $= 12$.

We know,
$$\frac{2b^2}{a} = 12 \Rightarrow b^2 = 6a \quad \text{…(1)}$$

Also,
$$c^2 = a^2 + b^2$$

$$16 = a^2 + 6a$$

$$a^2 + 6a – 16 = 0$$

Solving,
$$(a + 8)(a – 2) = 0$$

Hence, $a = 2$ (since $a > 0$).
Substituting in (1):
$$b^2 = 6(2) = 12$$

Thus,
$$a^2 = 4, \quad b^2 = 12$$

The equation is
$$\frac{x^2}{4} – \frac{y^2}{12} = 1$$

NCERT Question 14 : Find the equation of the hyperbola with vertices $(\pm 7, 0)$ and eccentricity $e = \frac{4}{3}$.

Solution:
Since the vertices lie on the x-axis, the equation of the hyperbola is of the form
$$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$$

Given, $a = 7$ and $e = \frac{4}{3}$.

We know,
$$e = \frac{c}{a} \Rightarrow c = ea = \frac{4}{3} \times 7 = \frac{28}{3}$$

Now,
$$b^2 = c^2 – a^2 = \frac{784}{9} – 49 = \frac{784 – 441}{9} = \frac{343}{9}$$

Hence,
$$a^2 = 49, \quad b^2 = \frac{343}{9}$$

The equation is
$$\frac{x^2}{49} – \frac{9y^2}{343} = 1$$

NCERT Question 15 : Find the equation of the hyperbola whose foci are $(0, \pm \sqrt{10})$ and which passes through the point $(2, 3)$.

Solution:
Since the foci lie on the y-axis, the equation of the hyperbola is of the form
$$\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$$

Given, $c = \sqrt{10}$ and the curve passes through $(2, 3)$.

We know,
$$b^2 = c^2 – a^2 = 10 – a^2 \quad \text{…(1)}$$

Substitute $(x, y) = (2, 3)$ in the equation:
$$\frac{3^2}{a^2} – \frac{2^2}{b^2} = 1$$

$$\frac{9}{a^2} – \frac{4}{10 – a^2} = 1$$

Simplify:
$$9(10 – a^2) – 4a^2 = a^2(10 – a^2)$$

$$90 – 9a^2 – 4a^2 = 10a^2 – a^4$$

$$a^4 – 23a^2 + 90 = 0$$

Factoring,
$$(a^2 – 18)(a^2 – 5) = 0$$

Since $a < c$, we take $a^2 = 5$.
From (1),
$$b^2 = 10 – 5 = 5$$

Hence, the equation is
$$\frac{y^2}{5} – \frac{x^2}{5} = 1$$

Learn complete NCERT Class 11 Maths Chapter on Hyperbola with step-by-step solutions and conceptual insights from Anand Classes — ideal for JEE, NDA, and CUET preparation.