Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Get insights from 88 questions on Maths NCERT Exemplar Solutions Class 12th Chapter Eleven, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Follow Ask Question
88

Questions

0

Discussions

3

Active Users

0

Followers

New question posted

2 months ago

0 Follower 1 View

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Since  (3,3) lies on x2a2-y2b2=1

9a2-9b2=1

Now, normal at  (3,3) is y-3=-a2b2 (x-3) ,

which passes through  (9,0)b2=2a2

So,  e2=1+b2a2=3

Also,  a2=92

(From (i) and (ii)

Thus,  a2, e2=92, 3

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

 x2a2+y2b2=1(ab);2b2a=10b2=5a

Now, ?(t)=512+t-t2=812-t-122
?(t)max=812=23=ee2=1-b2a2=49

a2=81 (From (i) and (ii)

So, a2+b2=81+45=126

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

(2x3+3xk)12

gen term =

=12Cr212r.3r.x363rrk

For constant term

36 – 3r – rk = 0

k=363rr

for r = 1, 2, 4

12Cr212-r>28

Possible values of k = 3, 1

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

y=2x2+x+2.... (i)

dydx=4x+1

Slope of normal

dxdy=14x+1

Equation of PQ y - β = 14α+1 (xα)

It passes (6, 4)

(4β) (4α+1)= (6α)

4α3+3α23α3=0

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

x 2 + y 2 2 x 4 y = 0

 Centre (1, 2) r = 5  

Equation of OQ is x . 0 + y . 0 – (x + 0) – 2 (y + 0) = 0

⇒ x + 2y = 0       ……… (i)

Equation of PQ is  x ( 1 + 5 ) + 2 y ( | x + 1 + 5 ) 2 ( y + 2 ) = 0

Solving (i) & (ii), Q ( 5 + 1 , 5 + 1 2 )  

= 1 2 | 6 + 2 5 + 4 5 = 4 2 | = 6 5 + 1 0 4 = 3 5 + 5 2

New answer posted

2 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

a 2 ( e 2 1 ) = b 2  

e = 5 2 b 2 = 3 a 2 2                

Length of latus rectum 2 b 2 a = 6 2  

3 a = 6 2 a = 2 2                

b = 2 3                

y = 2x + c is tangent to hyperbola

c 2 = a 2 m 2 b 2 = 2 0        

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

(a, 4a, 7). (3, 1, 2b)= 0

3a + 4a – 14b = 0 a – 2b = 0 ……. (i)

(a, 4a, 7) . (b, a, 2) = 0

ab – 4a2 + 14 = 0……… (ii)

(i) & (ii) =>2b2 – 16b2+ 14 = 0

b2=1a=2b=±2

Plane : x – y + z = 0

P (4, 5, 1)

4 + 5 + 1 = 10

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

A (1, 4, 3)

2x + my + nz = 4

M4+7m2+3n2=4

7m + 3n = 16

AM=(1,12,32)

21=m12=n32

m = 1, n = 3

Plane : 2x + y + 3z = 4

cosθ=|6112|2614=7291

AM = |2+4+94|14=714

cosθ=AMABAB=7/147/291=29114=213*72*7=

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

[x]2+2 [x+2]-7=0

[x]2+2 [x]+4-7=0 [x]=1, -3x [1,2) [-3, -2)

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Share Your College Life Experience

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.