Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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2 months ago

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A
alok kumar singh

Contributor-Level 10

Equation of tangent of slope m to y = x2 is y = mx 1 4 m 2 - …………. (i)

Equation of tangent of slope m to y = - (x - 2)2 is y = m (x – 2) + 1 4 m 2  …………… (ii)

If both equation represent the same line therefore on comparing (i) and (ii) we get m = 0, 4

therefore equation of tangent is y = 4x – 4

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2 months ago

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A
alok kumar singh

Contributor-Level 10

 x2 (52)2+y2 (53)2=1

Equation of tangent having slope m is

y=mx±53m2+53, which passes through (1, 3) and we get m1 + m2 = -4 and m1m2 = 449

Acute angle between the tangents is α  = tan-1 |m1m21+m1m2|=tan1 (2475)

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Vishal Baghel

Contributor-Level 10

 y2=x2

y=mx18m

This tangent pass through (2, 0)

m=±14i.e., onetangentisx4y2=0, 17r=9

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Ellipse is x24+y24=1e=12S (0, 2)

Chord of contact is

x2+ (222)y4=1

P (1, 2)Q2, 0

(SP)2+ (SQ)2=13

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 OPOA=tan15°

OA=OPcot15°

OPOC=tan45°OP=OC

Now, OP = OA282

OP2= (OP)2cot215°64

OP=323 (23)

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

Normal of plane P : = |i^j^k^213122|=4i^j3k^

Equation of plane P which passes through (2, 2) is 4x – y – 3z – 12 = 0

Now, A (3, 0, 0), B (0, 12 0), C (0, 4)

α=3, β12, γ=4P=α+β+γ=13

Now, volume of tetrahedron OABC

V=|16OA. (OB*OC)|=24

(V, P) = (24, 13)

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

Given,  L1:x1λ=y21=z32

and L2:x+262=y+183=z+28λ

are coplanar

|272031λ1223λ|=0

λ=3

Now, normal of plane P, which contains L1 and L2

equation of required plane P : 3x + 13y – 11z + 4 = 0

(0, 4, 5) does not lie on plane P.

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Vishal Baghel

Contributor-Level 10

H : x 2 a 2 y 2 b 2 = 1

Foci : S (ae, 0), S' (ae, 0)

Focus of parabola is (ae, 0)

Now, semi latus rectum of parabola = |SS'| = 2ae

Given, 4ae=e (2b2a)

B2 = 2a2……… (i)

Given, (22, 22) lies on H

1a21b2=18........ (ii)

From (i) and (ii)

a2=4, b2=8

? b2=a2 (e21)

e=3

equation of parabola is  y2 =83x

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

C : (x – 2)2 + y2 = 1

Equation of chord AB : 2x = 3

OA=OB=3

AM=32

AreaofΔOAB=12 (2AM) (OM)

=334sq.unit

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

D=14-1315ab6=021a-8b-66=0P:2x-3y+6z=15

so required distance =217=3

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