Rbse Solutions for Class 11 maths Chapter 10 Exercise 10.4 | Hyperbola

Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 10 Exercise 10.4 This exercise focuses on identifying the key properties of hyperbolas from their equations and finding the equation given specific conditions. All hyperbolas here are centered at the origin $(0, 0)$.

The key relation for any hyperbola is $c^2 = a^2 + b^2$, where $a$ is the semi-transverse axis, $b$ is the semi-conjugate axis, and $c$ is the distance from the center to the focus. Eccentricity is $e = c/a$.

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Finding Properties from the Equation (Exercises 1–6)

The two standard forms are:

1. Horizontal Transverse Axis: $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$ (Transverse axis is along the $x$-axis)
2. Vertical Transverse Axis: $\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$ (Transverse axis is along the $y$-axis)

Note: For a hyperbola, $a^2$ is always the denominator of the positive term.

1. $\frac{x^2}{16} – \frac{y^2}{9} = 1$

• Form: Horizontal (Transverse axis on $x$-axis).
• $a^2 = 16 \implies \mathbf{a = 4}$. $b^2 = 9 \implies \mathbf{b = 3}$.
• $c$: $c^2 = a^2 + b^2 = 16 + 9 = 25 \implies \mathbf{c = 5}$.
• Foci: $(\pm c, 0) = \mathbf{(\pm 5, 0)}$.
• Vertices: $(\pm a, 0) = \mathbf{(\pm 4, 0)}$.
• Eccentricity: $e = c/a = \mathbf{5/4}$.
• Latus Rectum Length: $\frac{2b^2}{a} = \frac{2(9)}{4} = \mathbf{9/2}$.

2. $\frac{y^2}{9} – \frac{x^2}{27} = 1$

• Form: Vertical (Transverse axis on $y$-axis).
• $a^2 = 9 \implies \mathbf{a = 3}$. $b^2 = 27 \implies \mathbf{b = \sqrt{27} = 3\sqrt{3}}$.
• $c$: $c^2 = a^2 + b^2 = 9 + 27 = 36 \implies \mathbf{c = 6}$.
• Foci: $(0, \pm c) = \mathbf{(0, \pm 6)}$.
• Vertices: $(0, \pm a) = \mathbf{(0, \pm 3)}$.
• Eccentricity: $e = c/a = \mathbf{6/3 = 2}$.
• Latus Rectum Length: $\frac{2b^2}{a} = \frac{2(27)}{3} = \mathbf{18}$.

3. $9y^2 – 4x^2 = 36$

Divide by 36: $\frac{9y^2}{36} – \frac{4x^2}{36} = 1 \implies \frac{y^2}{4} – \frac{x^2}{9} = 1$.

• Form: Vertical (Transverse axis on $y$-axis).
• $a^2 = 4 \implies \mathbf{a = 2}$. $b^2 = 9 \implies \mathbf{b = 3}$.
• $c$: $c^2 = 4 + 9 = 13 \implies \mathbf{c = \sqrt{13}}$.
• Foci: $(0, \pm c) = \mathbf{(0, \pm \sqrt{13})}$.
• Vertices: $(0, \pm a) = \mathbf{(0, \pm 2)}$.
• Eccentricity: $e = c/a = \mathbf{\sqrt{13}/2}$.
• Latus Rectum Length: $\frac{2b^2}{a} = \frac{2(9)}{2} = \mathbf{9}$.

4. $16x^2 – 9y^2 = 576$

Divide by 576: $\frac{16x^2}{576} – \frac{9y^2}{576} = 1 \implies \frac{x^2}{36} – \frac{y^2}{64} = 1$.

• Form: Horizontal (Transverse axis on $x$-axis).
• $a^2 = 36 \implies \mathbf{a = 6}$. $b^2 = 64 \implies \mathbf{b = 8}$.
• $c$: $c^2 = 36 + 64 = 100 \implies \mathbf{c = 10}$.
• Foci: $(\pm c, 0) = \mathbf{(\pm 10, 0)}$.
• Vertices: $(\pm a, 0) = \mathbf{(\pm 6, 0)}$.
• Eccentricity: $e = c/a = \mathbf{10/6 = 5/3}$.
• Latus Rectum Length: $\frac{2b^2}{a} = \frac{2(64)}{6} = \frac{64}{3}$.

5. $5y^2 – 9x^2 = 36$

Divide by 36: $\frac{5y^2}{36} – \frac{9x^2}{36} = 1 \implies \frac{y^2}{36/5} – \frac{x^2}{4} = 1$.

• Form: Vertical (Transverse axis on $y$-axis).
• $a^2 = 36/5 \implies \mathbf{a = 6/\sqrt{5} = 6\sqrt{5}/5}$. $b^2 = 4 \implies \mathbf{b = 2}$.
• $c$: $c^2 = a^2 + b^2 = \frac{36}{5} + 4 = \frac{36 + 20}{5} = \frac{56}{5} \implies \mathbf{c = \sqrt{56/5} = 2\sqrt{14/5}}$.
• Foci: $(0, \pm c) = \mathbf{(0, \pm \sqrt{56/5})}$.
• Vertices: $(0, \pm a) = \mathbf{(0, \pm 6/\sqrt{5})}$.
• Eccentricity: $e = c/a = \frac{\sqrt{56/5}}{6/\sqrt{5}} = \frac{\sqrt{56}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{6} = \frac{\sqrt{56}}{6} = \frac{2\sqrt{14}}{6} = \mathbf{\frac{\sqrt{14}}{3}}$.
• Latus Rectum Length: $\frac{2b^2}{a} = \frac{2(4)}{6/\sqrt{5}} = 8 \cdot \frac{\sqrt{5}}{6} = \mathbf{4\sqrt{5}/3}$.

6. $49y^2 – 16x^2 = 784$

Divide by 784: $\frac{49y^2}{784} – \frac{16x^2}{784} = 1 \implies \frac{y^2}{16} – \frac{x^2}{49} = 1$.

• Form: Vertical (Transverse axis on $y$-axis).
• $a^2 = 16 \implies \mathbf{a = 4}$. $b^2 = 49 \implies \mathbf{b = 7}$.
• $c$: $c^2 = 16 + 49 = 65 \implies \mathbf{c = \sqrt{65}}$.
• Foci: $(0, \pm c) = \mathbf{(0, \pm \sqrt{65})}$.
• Vertices: $(0, \pm a) = \mathbf{(0, \pm 4)}$.
• Eccentricity: $e = c/a = \mathbf{\sqrt{65}/4}$.
• Latus Rectum Length: $\frac{2b^2}{a} = \frac{2(49)}{4} = \mathbf{49/2}$.

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Finding the Equation of the Hyperbola (Exercises 7–15)

The equation is $\frac{x^2}{A} – \frac{y^2}{B} = 1$ or $\frac{y^2}{A} – \frac{x^2}{B} = 1$.

7. Vertices $(\pm 2, 0)$, foci $(\pm 3, 0)$.

1. Form: Vertices are on the $x$-axis $\implies$ Horizontal: $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$.
2. $a$ and $c$: Vertices $(\pm a, 0) \implies \mathbf{a = 2} \implies a^2 = 4$. Foci $(\pm c, 0) \implies \mathbf{c = 3}$.
3. $b^2$: $b^2 = c^2 – a^2 = 3^2 – 2^2 = 9 – 4 = \mathbf{5}$.$$\mathbf{\frac{x^2}{4} – \frac{y^2}{5} = 1}$$

8. Vertices $(0, \pm 5)$, foci $(0, \pm 8)$.

1. Form: Vertices are on the $y$-axis $\implies$ Vertical: $\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$.
2. $a$ and $c$: Vertices $(0, \pm a) \implies \mathbf{a = 5} \implies a^2 = 25$. Foci $(0, \pm c) \implies \mathbf{c = 8}$.
3. $b^2$: $b^2 = c^2 – a^2 = 8^2 – 5^2 = 64 – 25 = \mathbf{39}$.$$\mathbf{\frac{y^2}{25} – \frac{x^2}{39} = 1}$$

9. Vertices $(0, \pm 3)$, foci $(0, \pm 5)$.

1. Form: Vertical: $\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$.
2. $a$ and $c$: $\mathbf{a = 3} \implies a^2 = 9$. $\mathbf{c = 5}$.
3. $b^2$: $b^2 = c^2 – a^2 = 5^2 – 3^2 = 25 – 9 = \mathbf{16}$.$$\mathbf{\frac{y^2}{9} – \frac{x^2}{16} = 1}$$

10. Foci $(\pm 5, 0)$, the transverse axis is of length 8.

1. Form: Foci are on the $x$-axis $\implies$ Horizontal: $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$.
2. $a$ and $c$: Transverse axis length $2a = 8 \implies \mathbf{a = 4} \implies a^2 = 16$. Foci $(\pm c, 0) \implies \mathbf{c = 5}$.
3. $b^2$: $b^2 = c^2 – a^2 = 5^2 – 4^2 = 25 – 16 = \mathbf{9}$.$$\mathbf{\frac{x^2}{16} – \frac{y^2}{9} = 1}$$

11. Foci $(0, \pm 13)$, the conjugate axis is of length 24.

1. Form: Foci are on the $y$-axis $\implies$ Vertical: $\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$.
2. $b$ and $c$: Conjugate axis length $2b = 24 \implies \mathbf{b = 12} \implies b^2 = 144$. Foci $(0, \pm c) \implies \mathbf{c = 13}$.
3. $a^2$: $a^2 = c^2 – b^2 = 13^2 – 12^2 = 169 – 144 = \mathbf{25}$.$$\mathbf{\frac{y^2}{25} – \frac{x^2}{144} = 1}$$

12. Foci $(\pm 3\sqrt{5}, 0)$, the latus rectum is of length 8.

1. Form: Foci are on the $x$-axis $\implies$ Horizontal: $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$.
2. $c$ and L.R.: $\mathbf{c = 3\sqrt{5}} \implies c^2 = 9(5) = 45$. L.R. $= \frac{2b^2}{a} = 8 \implies 2b^2 = 8a \implies \mathbf{b^2 = 4a}$.
3. $a^2$: Substitute $b^2 = 4a$ into $c^2 = a^2 + b^2$:$$45 = a^2 + 4a$$$$a^2 + 4a – 45 = 0$$$$(a + 9)(a – 5) = 0$$Since $a$ must be positive, $\mathbf{a = 5} \implies a^2 = 25$.
4. $b^2$: $b^2 = 4a = 4(5) = \mathbf{20}$.$$\mathbf{\frac{x^2}{25} – \frac{y^2}{20} = 1}$$

13. Foci $(\pm 4, 0)$, the latus rectum is of length 12.

1. Form: Horizontal: $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$.
2. $c$ and L.R.: $\mathbf{c = 4} \implies c^2 = 16$. L.R. $= \frac{2b^2}{a} = 12 \implies 2b^2 = 12a \implies \mathbf{b^2 = 6a}$.
3. $a^2$: Substitute $b^2 = 6a$ into $c^2 = a^2 + b^2$:$$16 = a^2 + 6a$$$$a^2 + 6a – 16 = 0$$$$(a + 8)(a – 2) = 0$$Since $a > 0$, $\mathbf{a = 2} \implies a^2 = 4$.
4. $b^2$: $b^2 = 6a = 6(2) = \mathbf{12}$.$$\mathbf{\frac{x^2}{4} – \frac{y^2}{12} = 1}$$

14. Vertices $(\pm 7, 0)$, $e = 4/3$.

1. Form: Vertices are on the $x$-axis $\implies$ Horizontal: $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$.
2. $a$ and $c$: Vertices $(\pm a, 0) \implies \mathbf{a = 7} \implies a^2 = 49$. Eccentricity $e = c/a = 4/3$.$$c = ae = 7 \cdot \frac{4}{3} = \frac{28}{3} \implies c^2 = \frac{784}{9}$$
3. $b^2$: $b^2 = c^2 – a^2 = \frac{784}{9} – 49$$$b^2 = \frac{784 – 49(9)}{9} = \frac{784 – 441}{9} = \mathbf{\frac{343}{9}}$$$$\mathbf{\frac{x^2}{49} – \frac{y^2}{343/9} = 1 \quad \text{or} \quad \frac{x^2}{49} – \frac{9y^2}{343} = 1}$$

15. Foci $(0, \pm \sqrt{10})$, passing through $(2, 3)$.

1. Form: Foci are on the $y$-axis $\implies$ Vertical: $\frac{y^2}{a^2} – \frac{x^2}{b^2} = 1$.
2. $c$ and $a^2/b^2$ relation: Foci $(0, \pm c) \implies \mathbf{c = \sqrt{10}} \implies c^2 = 10$.$$c^2 = a^2 + b^2 \implies \mathbf{10 = a^2 + b^2} \quad \text{(i)}$$
3. Use the point $(2, 3)$:$$\frac{3^2}{a^2} – \frac{2^2}{b^2} = 1 \implies \mathbf{\frac{9}{a^2} – \frac{4}{b^2} = 1} \quad \text{(ii)}$$
4. Solve the system: From (i), $b^2 = 10 – a^2$. Substitute into (ii):$$\frac{9}{a^2} – \frac{4}{10 – a^2} = 1$$Multiply by $a^2(10 – a^2)$:$$9(10 – a^2) – 4a^2 = a^2(10 – a^2)$$$$90 – 9a^2 – 4a^2 = 10a^2 – a^4$$$$a^4 – 23a^2 + 90 = 0$$
5. Solve the quadratic for $a^2$: Let $u = a^2$.$$u^2 – 23u + 90 = 0$$$$(u – 18)(u – 5) = 0$$$$u = 18 \quad \text{or} \quad u = 5$$
   • Case 1: $a^2 = 18$. Then $b^2 = 10 – 18 = -8$. (Impossible since $b^2 > 0$).
   • Case 2: $\mathbf{a^2 = 5}$. Then $b^2 = 10 – 5 = \mathbf{5}$.
6. Final Equation:$$\mathbf{\frac{y^2}{5} – \frac{x^2}{5} = 1 \quad \text{or} \quad y^2 – x^2 = 5}$$

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