NCERT Solutions Class 11 Maths Chapter 10 Conic Sections Exercise 10.4

Exercise 10.4

1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1$

Ans: The given equation is $\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1$ or $\dfrac{{{x}^{2}}}{{{4}^{2}}}-\dfrac{{{y}^{2}}}{{{3}^{2}}}=1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=4$ and $b=3$. 

We know that ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$. 

$\begin{align}  & \therefore {{c}^{2}}={{4}^{2}}+{{3}^{2}} \\  & \Rightarrow {{c}^{2}}=16+9 \\  & \Rightarrow {{c}^{2}}=25 \\  & \Rightarrow c=\sqrt{25}=5 \\ \end{align}$

Therefore, 

The coordinates of the foci are $\left( \pm 5,0 \right)$ 

The coordinates of the vertices are $(\pm 4,0)$ 

Eccentricity, $e=\dfrac{c}{a}=\dfrac{5}{4}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 9}{4}=\dfrac{9}{2}$

2. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{27}=1$

Ans: The given equation is $\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{27}=1$ or $\dfrac{{{y}^{2}}}{{{3}^{2}}}-\dfrac{{{x}^{2}}}{{{\left( \sqrt{27} \right)}^{2}}}=1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=3$ and $b=\sqrt{27}$. 

We know that ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$. 

$\begin{align}  & \therefore {{c}^{2}}={{3}^{2}}+{{\left( \sqrt{27} \right)}^{2}} \\  & \Rightarrow {{c}^{2}}=9+27 \\ & \Rightarrow {{c}^{2}}=36 \\  & \Rightarrow c=\sqrt{36}=6 \\ \end{align}$

Therefore, 

The coordinates of the foci are $\left( 0,\pm 6 \right)$ 

The coordinates of the vertices are $(0,\pm 3)$ 

Eccentricity, $e=\dfrac{c}{a}=\dfrac{6}{3}=2$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 27}{3}=18$

3. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $9{{y}^{2}}-4{{x}^{2}}=36$

Ans: The given equation is $9{{y}^{2}}-4{{x}^{2}}=36$. 

It can be written as 

$9{{y}^{2}}-4{{x}^{2}}=36$

Or, $\dfrac{{{y}^{2}}}{4}-\dfrac{{{x}^{2}}}{9}=1$ 

Or, $\dfrac{{{y}^{2}}}{{{2}^{2}}}-\dfrac{{{x}^{2}}}{{{3}^{2}}}=1$ ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=2$ and $b=3$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{c}^{2}}=4+9 \\  & \Rightarrow {{c}^{2}}=13 \\  & \Rightarrow c=\sqrt{13} \\ \end{align}$

Therefore, 

The coordinates of the foci are $\left( 0,\pm \sqrt{13} \right)$ 

The coordinates of the vertices are $(0,\pm 2)$ 

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{2}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 9}{2}=9$

4. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $16{{x}^{2}}-9{{y}^{2}}=576$

Ans: The given equation is $16{{x}^{2}}-9{{y}^{2}}=576$. 

It can be written as 

$16{{x}^{2}}-9{{y}^{2}}=576$

Or, $\dfrac{{{x}^{2}}}{36}-\dfrac{{{y}^{2}}}{64}=1$ 

Or, $\dfrac{{{x}^{2}}}{{{6}^{2}}}-\dfrac{{{y}^{2}}}{{{8}^{2}}}=1$ ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=6$ and $b=8$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{c}^{2}}=36+64 \\  & \Rightarrow {{c}^{2}}=100 \\  & \Rightarrow c=\sqrt{100} \\  & \Rightarrow c=10 \\ \end{align}$

Therefore, 

The coordinates of the foci are $\left( \pm \sqrt{10},0 \right)$ 

The coordinates of the vertices are $(\pm 6,0)$ 

Eccentricity, $e=\dfrac{c}{a}=\dfrac{10}{6}=\dfrac{5}{3}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 64}{6}=\dfrac{64}{3}$

5. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $5{{y}^{2}}-9{{x}^{2}}=36$

Ans: The given equation is $5{{y}^{2}}-9{{x}^{2}}=36$. 

$\Rightarrow \dfrac{{{y}^{2}}}{\dfrac{36}{5}}-\dfrac{{{x}^{2}}}{4}=1$ 

$\dfrac{{{y}^{2}}}{\left( \dfrac{6}{\sqrt{5}} \right)^2}-\dfrac{{{x}^{2}}}{{{2}^{2}}}=1$ ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=\dfrac{6}{\sqrt{5}}$ and $b=2$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{c}^{2}}=\dfrac{36}{5}+4 \\  & \Rightarrow {{c}^{2}}=\dfrac{56}{5} \\  & \Rightarrow c=\sqrt{\dfrac{56}{5}} \\  & \Rightarrow c=\dfrac{2\sqrt{14}}{\sqrt{5}} \\ \end{align}$

Therefore, 

The coordinates of the foci are $\left( 0,\pm \dfrac{2\sqrt{14}}{\sqrt{5}} \right)$ 

The coordinates of the vertices are $\left( 0,\pm \dfrac{6}{\sqrt{5}} \right)$ 

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\left( \dfrac{2\sqrt{14}}{\sqrt{5}} \right)}{\left( \dfrac{6}{\sqrt{5}} \right)}=\dfrac{\sqrt{14}}{3}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 4}{\left( \dfrac{6}{\sqrt{5}} \right)}=\dfrac{4\sqrt{5}}{3}$

6. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $49{{y}^{2}}-16{{x}^{2}}=784$

Ans: The given equation is $49{{y}^{2}}-16{{x}^{2}}=784$. 

It can be written as 

$49{{y}^{2}}-16{{x}^{2}}=784$

Or, $\dfrac{{{y}^{2}}}{16}-\dfrac{{{x}^{2}}}{49}=1$ 

Or, $\dfrac{{{y}^{2}}}{{{4}^{2}}}-\dfrac{{{x}^{2}}}{{{7}^{2}}}=1$ ………(1)

On comparing equation (1) with the standard equation of hyperbola i.e., $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$, we’ll get $a=4$ and $b=7$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{c}^{2}}=16+49 \\  & \Rightarrow {{c}^{2}}=65 \\  & \Rightarrow c=\sqrt{65} \\ \end{align}$

Therefore, 

The coordinates of the foci are $\left( 0,\pm \sqrt{65} \right)$ 

The coordinates of the vertices are $(0,\pm 4)$ 

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{65}}{4}$

Length of latus rectum $=\dfrac{2{{b}^{2}}}{a}=\dfrac{2\times 49}{4}=\dfrac{49}{2}$

7. Find the equation of the hyperbola satisfying the give conditions: Vertices $(\pm 2,0)$, foci $(\pm 3,0)$ 

Ans: Vertices $(\pm 2,0)$, foci $(\pm 3,0)$

Here, the vertices are on the x-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Since the vertices are $(\pm 2,0)$, $a=2$. 

Since the foci are $(\pm 3,0)$, $c=3$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{2}^{2}}+{{b}^{2}}={{3}^{2}} \\  & \Rightarrow {{b}^{2}}=9-4 \\  & \Rightarrow {{b}^{2}}=5 \\ \end{align}$

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{4}-\dfrac{{{y}^{2}}}{5}=1$

8. Find the equation of the hyperbola satisfying the give conditions: Vertices $(0,\pm 5)$, foci $(0,\pm 8)$

Ans: Vertices $(0,\pm 5)$, foci $(0,\pm 8)$

Here, the vertices are on the y-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$

Since the vertices are $(0,\pm 5)$, $a=5$. 

Since the foci are $(0,\pm 8)$, $c=8$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{5}^{2}}+{{b}^{2}}={{8}^{2}} \\  & \Rightarrow {{b}^{2}}=64-25 \\  & \Rightarrow {{b}^{2}}=39 \\ \end{align}$

Thus, the equation of the hyperbola is $\dfrac{{{y}^{2}}}{25}-\dfrac{{{x}^{2}}}{39}=1$

9. Find the equation of the hyperbola satisfying the give conditions: Vertices $(0,\pm 3)$, foci $(0,\pm 5)$ 

Ans: Vertices $(0,\pm 3)$, foci $(0,\pm 5)$

Here, the vertices are on the y-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$

Since the vertices are $(0,\pm 3)$, $a=3$. 

Since the foci are $(0,\pm 5)$, $c=5$.

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}   & \therefore {{3}^{2}}+{{b}^{2}}=25 \\  & \Rightarrow {{b}^{2}}=25-9 \\  & \Rightarrow {{b}^{2}}=16 \\ \end{align}$

Thus, the equation of the hyperbola is $\dfrac{{{y}^{2}}}{9}-\dfrac{{{x}^{2}}}{16}=1$

10. Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 5,0)$, the transverse axis is of length $8$. 

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

Here, the foci are on the x-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. 

Since the foci are $(\pm 5,0)$, $c=5$. 

Since the length of the transverse axis is $8$, $2a=8\Rightarrow a=4$. 

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{4}^{2}}+{{b}^{2}}=25 \\  & \Rightarrow {{b}^{2}}=25-16 \\  & \Rightarrow {{b}^{2}}=9 \\ \end{align}$

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1$

11. Find the equation of the hyperbola satisfying the give conditions: Foci $(0,\pm 13)$, the conjugate axis is of length $24$. 

Ans: Foci $(0,\pm 13)$, the transverse axis is of length $24$. 

Here, the foci are on the y-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$. 

Since the foci are $(0,\pm 13)$, $c=13$. 

Since the length of the transverse axis is $24$, $2b=24\Rightarrow b=12$. 

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{a}^{2}}+{{12}^{2}}={{13}^{2}} \\  & \Rightarrow {{a}^{2}}=169-144 \\  & \Rightarrow {{a}^{2}}=25 \\ \end{align}$

Thus, the equation of the hyperbola is $\dfrac{{{y}^{2}}}{25}-\dfrac{{{x}^{2}}}{144}=1$

12. Find the equation of the hyperbola satisfying the give conditions: Foci $\left( \pm 3\sqrt{5},0 \right)$, the latus rectum is of length $8$. 

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

Here, the foci are on the x-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Since the foci are $\left( \pm 3\sqrt{5},0 \right)$, $c=\pm 3\sqrt{5}$

Length of latus rectum = $8$

$\begin{align}  & \Rightarrow \dfrac{2{{b}^{2}}}{a}=8 \\  & \Rightarrow {{b}^{2}}=4a \\ \end{align}$

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{a}^{2}}+4a=45 \\  & \Rightarrow {{a}^{2}}+4a-45=0 \\  & \Rightarrow {{a}^{2}}+9a-5a-45=0 \\  & \Rightarrow \left( a+9 \right)\left( a-5 \right)=0 \\  & \Rightarrow a=-9,5 \\ \end{align}$

Since a is non-negative, $a=5$. 

$\therefore {{b}^{2}}=4a=4\times 5=20$

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{20}=1$

13. Find the equation of the hyperbola satisfying the give conditions: Foci $\left( \pm 4,0 \right)$, the latus rectum is of length $12$. 

Ans: Foci $\left( \pm 4,0 \right)$, the latus rectum is of length $12$. 

Here, the foci are on the x-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Since the foci are $\left( \pm 4,0 \right)$, $c=4$. 

Length of latus rectum = $12$

$\begin{align}  & \Rightarrow \dfrac{2{{b}^{2}}}{a}=12 \\  & \Rightarrow {{b}^{2}}=6a \\ \end{align}$

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{a}^{2}}+6a=16 \\ & \Rightarrow {{a}^{2}}+6a-16=0 \\  & \Rightarrow {{a}^{2}}+8a-2a-16=0 \\  & \Rightarrow \left( a+8 \right)\left( a-2 \right)=0 \\ & \Rightarrow a=-8,2 \\ \end{align}$

Since a is non-negative, $a=2$. 

$\therefore {{b}^{2}}=6a=6\times 2=12$

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{4}-\dfrac{{{y}^{2}}}{12}=1$

14. Find the equation of the hyperbola satisfying the give conditions: Vertices $\left( \pm 7,0 \right)$, $e=\dfrac{4}{3}$

Ans: Vertices $\left( \pm 7,0 \right)$, $e=\dfrac{4}{3}$ 

Here, the vertices are on the x-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$

Since the vertices are $\left( \pm 7,0 \right)$, $a=7$. 

It is given that $e=\dfrac{4}{3}$

$\therefore \dfrac{c}{a}=\dfrac{4}{3}$ $\left[ e=\dfrac{c}{a} \right]$

$\begin{align}  & \Rightarrow \dfrac{c}{7}=\dfrac{4}{3} \\  & \Rightarrow c=\dfrac{28}{3} \\ \end{align}$

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\begin{align}  & \therefore {{7}^{2}}+{{b}^{2}}={{\left( \dfrac{28}{3} \right)}^{2}} \\  & \Rightarrow {{b}^{2}}=\dfrac{784}{9}-49 \\  & \Rightarrow {{b}^{2}}=\dfrac{784-441}{9}=\dfrac{343}{9} \\ \end{align}$

Thus, the equation of the hyperbola is $\dfrac{{{x}^{2}}}{49}-\dfrac{{9{y}^{2}}}{343}=1$

15. Find the equation of the hyperbola satisfying the given conditions: Foci $\left( 0,\pm \sqrt{10} \right)$, passing through $\left( 2,3 \right)$

Ans: Foci $\left( 0,\pm \sqrt{10} \right)$, passing through $\left( 2,3 \right)$ 

Here, the foci are on the y-axis. 

Therefore, the equation of the hyperbola is of the form $\dfrac{{{y}^{2}}}{{{a}^{2}}}-\dfrac{{{x}^{2}}}{{{b}^{2}}}=1$ .

Since the foci are $\left( 0,\pm \sqrt{10} \right)$, $c=\sqrt{10}$ . 

We know that ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\therefore {{a}^{2}}+{{b}^{2}}=10$

${{a}^{2}}+{{b}^{2}}={{c}^{2}}$. 

$\Rightarrow {{b}^{2}}=10-{{a}^{2}}$ … (1) 

Since the hyperbola passes through point $\left( 2,3 \right)$, 

$\dfrac{9}{{{a}^{2}}}-\dfrac{4}{{{b}^{2}}}=1$ ….(2) 

From equations (1) and (2), we’ll get  

$\begin{align}  & \dfrac{9}{{{a}^{2}}}-\dfrac{4}{{{\left( 10-a^{2}\right)}}}=1 \\  & \Rightarrow 9\left( 10-{{a}^{2}} \right)-4{{a}^{2}}={{a}^{2}}\left( 10-{{a}^{2}} \right) \\  & \Rightarrow 90-9{{a}^{2}}-4{{a}^{2}}=10{{a}^{2}}-{{a}^{4}} \\  & \Rightarrow {{a}^{4}}-10{{a}^{2}}-9{{a}^{2}}-4{{a}^{2}}+90=0 \\  & \Rightarrow {{a}^{4}}-23{{a}^{2}}+90=0 \\  & \Rightarrow {{a}^{4}}-18{{a}^{2}}-5{{a}^{2}}+90=0 \\  & \Rightarrow {{a}^{2}}({{a}^{2}}-18)-5({{a}^{2}}-18)=0 \\  & \Rightarrow ({{a}^{2}}-5)({{a}^{2}}-18)=0 \\  & \Rightarrow {{a}^{2}}=18,5 \\ \end{align}$

In hyperbola, $c>a$ i.e., ${{c}^{2}}>{{a}^{2}}$

$\begin{align}  & \therefore a{}^{2}=5 \\  & \Rightarrow {{b}^{2}}=10-{{a}^{2}} \\  & \Rightarrow {{b}^{2}}=10-5=5 \\ \end{align}$

Thus, the equation of the hyperbola is $\dfrac{{{y}^{2}}}{5}-\dfrac{{{x}^{2}}}{5}=1$

Class 11 Maths Chapter 10: Exercises Breakdown

Conclusion

NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections: Exercise 10.4 provide a comprehensive understanding of hyperbolas. These solutions cover important concepts like the standard equation, eccentricity, foci, and asymptotes, making it easier for students to grasp the properties of hyperbolas and solve related problems confidently. By following these step-by-step solutions, students can strengthen their conceptual understanding and perform well in exams. Don't forget to download the FREE PDF for easy access and efficient exam preparation.

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