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NCERT Solutions for Class 11 Maths Chapter 10: Conic Sections - Exercise 10.4

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NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.4 (Ex 10.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 10 Conic Sections Exercise 10.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 10 - Conic Sections

Exercise:

Exercise - 10.4

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

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Access NCERT Solutions for Class 11 Maths Chapter 10- Conic Sections

Exercise 10.4

Refer the page 23 to 39 for the exercise 10.4 in the PDF

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 obtain \[a = 4\] and \[b = 3\] , 

We know that 

\[{c^2} = {a^2} + {b^2}\]

Also, we know that \[a = 4\]  and \[b = 3\].

$\therefore {c^2} = {4^2} + {3^2}$

Put the value of $b$ and $a$ to determine the value of $c$.

${c^2} = 25$

$c = \sqrt {25} $

$c = 5$

Therefore,

The coordinates of the foci are \[\left( { \pm 5,0} \right)\].

The coordinates of the vertices are \[\left( { \pm 4,0} \right)\]

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{5}{4}$

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

So, $\dfrac{{2{b^2}}}{a} = \dfrac{9}{2}$

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 obtain \[a = 3\]and\[b = \sqrt {27} \]. 

We know that 

\[{c^2} = {b^2} + {a^2}\]

Put the value of $b$ and $a$ to determine the value of $c$.

$\therefore {c^2} = {3^2} + {\left( {\sqrt {27} } \right)^2}$

${c^2} = 9 + 27$

${c^2} = 36$

$c = \sqrt {36} $

$c = 6$

Therefore,

The coordinates of the foci are \[\left( {0, \pm 6} \right)\].

The coordinates of the vertices are \[\left( {0, \pm 3} \right)\]

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{6}{3}$

$e = 2$

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 

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

Dividing 36 in both side

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

$\dfrac{{{y^2}}}{{{2^2}}} - \dfrac{{{x^2}}}{{{3^2}}} = 1$

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

We know that,

\[{c^2} = {b^2} + {a^2}\]

$\therefore {c^2} = 4 + 9$

${c^2} = 13$

$c= \sqrt {13} $

Therefore,

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

The coordinates of the vertices are \[\left( {0, \pm 2} \right)\]

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{{\sqrt {13} }}{2}$

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 

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

Dividing by 576 in both side

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

$\dfrac{{{x^2}}}{{{6^2}}} - \dfrac{{{y^2}}}{{{8^2}}} = 1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$, we obtain \[a = 6\]and\[b = 8\] , 

We know that 

\[{c^2} = {b^2} + {a^2}\]

Put the value of $b$ and $a$ to determine the value of $c$.

$\therefore {c^2} = 36 + 64$

${c^2} = 100$

$c = 10$

Therefore,

The coordinates of the foci are \[\left( { \pm 10,0} \right)\].

The coordinates of the vertices are \[\left( { \pm 6,0} \right)\].

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{{10}}{6}$

$\dfrac{c}{a} = \dfrac{5}{3}$

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

So, $\dfrac{{2{b^2}}}{a} = \dfrac{{64}}{3}$

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 $5{y^2} - 9{x^2} = 36$. 

It can be written as 

$5{y^2} - 9{x^2} = 36$

Dividing 36 in both side and arrange the question similar with $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$

$\dfrac{{{y^2}}}{{\left( {\dfrac{{36}}{5}} \right)}} - \dfrac{{{x^2}}}{4} = 1$

$\dfrac{{{y^2}}}{{{{\left( {\dfrac{6}{{\sqrt 5 }}} \right)}^2}}} - \dfrac{{{x^2}}}{{{2^2}}} = 1$………………………(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 obtain \[a = \dfrac{6}{{\sqrt 5 }}\] and\[b = 2\] ,

We know that 

\[{a^2} + {b^2} = {c^2}\]

Put the value of $b$ and $a$, also simplify the equation to determine the value of $c$.

$\therefore {c^2} = \dfrac{{36}}{5} + 4$

${c^2} = \dfrac{{36 + 20}}{5}$

${c^2} = \dfrac{{56}}{5}$

$c=\pm \sqrt{\frac{56}{5}}$

$c= 2\sqrt{\frac{14}{5}}$

Therefore,

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

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

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{{\left( {\dfrac{{2\sqrt {14} }}{{\sqrt 5 }}} \right)}}{{\left( {\dfrac{6}{{\sqrt 5 }}} \right)}}$

$e = \dfrac{{2\sqrt {14} }}{6}$

$e = \dfrac{{\sqrt {14} }}{3}$

Length of latus rectum, $\dfrac{{2{b^2}}}{a} = \dfrac{{2 \times 4}}{{\dfrac{6}{{\sqrt 5 }}}} = \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 

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

By dividing 784 in both sides, we get,

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

$\dfrac{{{y^2}}}{{{4^2}}} - \dfrac{{{x^2}}}{{{7^2}}} = 1$

On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$, we obtain \[a = 4\] and \[b = 7\]. 

We know that 

\[{c^2} = {b^2} + {a^2}\]

Put the value of $b$ and $a$ and simplify the equation to determine the value of $c$.

$\therefore {c^2} = 16 + 49$

${c^2} = 65$

$c= \sqrt{65}$

Therefore,

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

The coordinates of the vertices are \[\left( {0, \pm 4} \right)\]

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

As we know the value of $c$ and $a$. Thus,

$e = \dfrac{{\sqrt {65} }}{4}$

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

So, $\dfrac{{2{b^2}}}{a} = \dfrac{{49}}{2}$.

7. Find the equation of the hyperbola satisfying the given conditions: Vertices \[\left( { \pm {\mathbf{2}},{\mathbf{0}}} \right),\]foci \[\left( { \pm {\mathbf{3}},{\mathbf{0}}} \right)\].

Ans: Vertices \[\left( { \pm {\mathbf{2}},{\mathbf{0}}} \right)\] , foci \[\left( { \pm {\mathbf{3}},{\mathbf{0}}} \right)\]

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 \[\left( { \pm 2,0} \right)\], so $a = 2$. 

Since the foci are \[\left( { \pm 3,0} \right)\], so $c = 3$. 

We know that

${a^2} + {b^2} = {c^2}$

Put the value of $a$ and $c$ and simplify the equation to determine the value of ${b^2}$.

$\therefore {2^2} + {b^2} = {3^2}$

\[{b^2} = 9 - 4\]

${b^2} = 5$

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 given conditions: Vertices \[\left( {{\mathbf{0}}, \pm {\mathbf{5}}} \right)\], foci \[\left( {{\mathbf{0}}, \pm {\mathbf{8}}} \right)\]

Ans:  Vertices \[\left( {{\mathbf{0}}, \pm {\mathbf{5}}} \right)\] , foci \[\left( {{\mathbf{0}}, \pm 8} \right)\]

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 \[\left( {{\mathbf{0}}, \pm {\mathbf{5}}} \right)\] , so $a = 5$. 

Since the foci are \[\left( {{\mathbf{0}}, \pm 8} \right)\] , so $c = 8$. 

We know that 

${a^2} + {b^2} = {c^2}$

Put the value of $a$ and $c$ to get the value of $b$,

$\therefore {5^2} + {b^2} = {8^2}$

${b^2} = {8^2} - {5^2}$

\[{b^2} = 64 - 25\]

${b^2} = 39$

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 given conditions: Vertices \[\left( {{\mathbf{0}}, \pm 3} \right)\] , foci \[\left( {{\mathbf{0}}, \pm 5} \right)\]

Ans:  Vertices \[\left( {{\mathbf{0}}, \pm 3} \right)\] , foci \[\left( {{\mathbf{0}}, \pm 5} \right)\]

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 \[\left( {{\mathbf{0}}, \pm 3} \right)\] , so $a = 3$. 

Since the foci are \[\left( {{\mathbf{0}}, \pm 5} \right)\] , so $c = 5$. 

We know that 

${a^2} + {b^2} = {c^2}$

Put the value of $c$ and $a$ and simplify the equation to determine the value of ${b^2}$.

$ \therefore {3^2} + {b^2} $ = 25

\[{b^2} = 25 - 9\]

${b^2} = 16$

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 given conditions: Foci \[\left( { \pm 5,0} \right)\] , the transverse axis is of length 8. 

Ans: Foci \[\left( { \pm 5,0} \right)\] , &  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 \[\left( { \pm 5,0} \right)\] , so $c = 5$. 

Since the length of the transverse axis is 8,

\[2a = 8\]

Dividing both side with 2

\[a = 4\]

We know that 

${a^2} + {b^2} = {c^2}$

Put the value of $c$ and $a$ and simplify the equation to determine the value of $b$.

$ \therefore {4^2} + {b^2} $ = 25

\[{b^2} = 25 - 16\]

${b^2} = 9$

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 given conditions: Foci \[\left( {{\mathbf{0}}, \pm 13} \right)\] , the conjugate axis is of length 24. 

Ans: Foci \[\left( {{\mathbf{0}}, \pm 13} \right)\] , & the conjugate 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 \[\left( {{\mathbf{0}}, \pm 13} \right)\] , so  $c = 13$. 

Since the length of the  conjugate axis  is 24,

\[2b = 24\]

Dividing 2 on both sides,

\[b = 12\]

We know that 

${a^2} + {b^2} = {c^2}$

Put the value of $b$ and $c$, also simplify the equation to determine the value of $a$.

\[\therefore {a^2} + {12^2} = {13^2}\]

\[{a^2} = 169 - 144\]

${a^2} = 25$

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 given 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 \[\left( { \pm 3\sqrt 5 ,0} \right)\] , so  $c =  \pm 3\sqrt 5 $. 

Since the length of the latus rectum is 8,

\[\dfrac{{2{b^2}}}{a} = 8\]

Multiplying both sides by $\dfrac{a}{2}$,

\[{b^2} = \dfrac{{8a}}{2}\]

\[{b^2} = 4a\]

We know that 

${a^2} + {b^2} = {c^2}$

Put the value of $b$ and $c$,  simplify the equation to determine the value of $a$.

\[\therefore {a^2} + 4a = 45\]

\[{a^2} + 4a - 45 = 0\]

\[{a^2} + 9a - 5a - 45 = 0\]

$\left( {a + 9} \right)\left( {a - 5} \right) = 0$

$a =  - 9, 5$

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 given 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 \[\left( { \pm 4,0} \right)\] , so  $c = 4$. 

Since the latus rectum is of length 12. 

\[\dfrac{{2{b^2}}}{a} = 12\]

Multiply both sides with $\dfrac{a}{2}$,

\[{b^2} = 6a\]…………….(1)

We know that 

${a^2} + {b^2} = {c^2}$…………..(2)

Merge the equation (1) and (2) and simplify the equation to get the value of $a$,

\[\therefore {a^2} + 6a = 16\]

\[{a^2} + 6a - 16 = 0\]

\[{a^2} + 8a - 2a - 16 = 0\]

\[\left( {a + 8} \right)\left( {a - 2} \right) = 0\]

\[a =  - 8, 2\]

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 given 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 \[\left( { \pm 7,0} \right)\] , so $a = 7$. 

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

$\therefore \dfrac{c}{a} = \dfrac{4}{3} $

$\dfrac{c}{7} = \dfrac{4}{3}$

Multiply both sides with 7

$c = \dfrac{{28}}{3}$

We know that 

${a^2} + {b^2} = {c^2}$

Put the value of $c$ and $a$ ,  also simplify the equation to determine the value of $b$.

\[\therefore {7^2} + {b^2} = {\left( {\dfrac{{28}}{3}} \right)^2}\]

${b^2} = \dfrac{{784}}{9} - 49$

${b^2} = \dfrac{{784 - 441}}{9}$

${b^2} = \dfrac{{343}}{9}$

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 \[\left( {{\mathbf{0}}, \pm \sqrt {10} } \right)\] , passing through \[\left( {{\mathbf{2}},{\mathbf{3}}} \right)\]

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

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 \[\left( {{\mathbf{0}}, \pm \sqrt {10} } \right)\] , so $c = \sqrt {10} $. 

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

Put the value of $c$,

\[\therefore {a^2} + {b^2} = 10\]

\[{b^2} = 10 - {a^2}\]………..… (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 obtain the value of ${a^2}$

$\dfrac{9}{{{a^2}}} - \dfrac{4}{{\left( {10 - {a^2}} \right)}} = 1$

$9(10 - {a^2}) - 4{a^2} = {a^2}\left( {10 - {a^2}} \right)$

$90 - 9{a^2} - 4{a^2} = 10{a^2} - {a^4}$

${a^4} - 23{a^2} + 90 = 0$

${a^4} - 18{a^2} - 5{a^2} + 90 = 0$

${a^2}\left( {{a^2} - 18} \right) - 5\left( {{a^2} - 18} \right) = 0$

\[{a^2} = 18,5\]

In hyperbola,

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

$\therefore {a^2} = 5$

Since the value of $a$ is positive as $e>1$ so ${b^2} = 10 - {a^2}$

As we know the value of $a$,

${b^2} = 10 - 5 = 5$

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

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Exercise 10.4

Opting for the NCERT solutions for Ex 10.4 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.4 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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FAQs on NCERT Solutions for Class 11 Maths Chapter 10: Conic Sections - Exercise 10.4

1. How many questions are there in NCERT Solutions for Class 11 Maths Chapter 10 Conic sections (Ex 10.4) Exercise 10.4?

There are a total of 15 questions in Exercise 10.4 of Chapter 10 Maths. In NCERT Conic sections deal with different shapes, circular cones, etc., and also consist of several sections where students get to learn about the semi-minor axis, the distance between focus and the central point of an ellipse, etc. You can refer to Vedantu in NCERT Solutions for Class 11 Maths Chapter 10 Conic sections (Ex 10.4) Exercise 10.4. Here accurate solutions are crafted by Subject experts that automatically bring excellent scores.

2. How can I revise exercise 10.4 in the chapter 10 Conic sections?

Revision is the most important part of any exam preparation.Experts always advise students to solve all the questions given in the NCERT math exercises many times. The more you practice, the more will be your accuracy and your understanding level. You can refer to Vedantu for solutions to all questions given in Chapter 10 Conic sections. Here you can also get revision notes for conic sections. To get all these free study materials, you can visit Vedantu.com or Vedantu mobile app.

3. What is the importance of Vedantu’s NCERT Solutions for Class 11 Maths Chapter 10 Conic sections (Ex 10.4) Exercise 10.4?

Ventantu, India’s top online educational platform provides NCERT Solutions for all subjects free of cost. Some of the major advantages are:

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4. How do I solve all the questions provided in Chapter 10 Conic sections Exercises 10.4?

Before you start solving exercise questions, try to understand the basic concepts, formulas, and also examples given in chapter 10. Then you can start to solve these questions and always try to do these questions by yourself. If you face any difficulties, you can refer to Vedantu for the solution to all the questions by subject experts. You can download all these solutions on the official website Vedantu (Vedantu.com) and Vedantu mobile app.

5. Do I need to practice all the questions provided in NCERT Solutions for Class 11 Maths Chapter 10 Conic sections (Ex 10.4) Exercise 10.4?

Definitely, you must practice all the questions provided in the NCERT textbook to score good marks. Class 11 Maths NCERT solutions can be considered the most useful resource as it contains different varieties of questions requiring proper concept and understanding to solve. With continuous practice, you can prepare yourself for any uncommon or tricky questions in your exams. Vedantu provides detailed step-by-step NCERT Solutions for all chapters in maths free of cost.