Conic Sections

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New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

 x2a2y21=1

Length of latus rectum = 2a

andx24+y23=1

length of latus rectum = 62 = 3

? 2a=3a=23

12 (eH2+eH2)=12 [ (1+94)+ (134)]=12 [134+14] = 12 * 144=42

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

OM2 = OP2 = PM2

| 1 + r 2 | = r 2 1

r = 3

equation of circle is  (x1)2+ (y3)2=32

h+k+r=7

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Required area

42 (4yy22)dy

(4yy22y36)42=18squnits

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

 x2a2+y24=1

Δ=12*a (1+cosθ).4sinθ

ΔmaxΔ=63a=4

e=32

 

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Equation of tangent at P (x, y) is Y = dydx (Xx)

A/q, 2xydxdy=02dyy=dxx

2lny=lnx+lncy2=xc

It passes through (3, 3), c = 3

y2=3x Length of latus rectum = 3

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

2 y e x / y 2 d x + ( y 2 4 x e x / y 2 ) d y = 0

  2 e 2 / y 2 ( y d x 2 x d y ) + y 2 d y = 0            

2 e x / y 2 d ( x y 2 ) + d y y = 0

Integrating   2 e x / y 2 + I n y = c

y = 1, n = 0  c = 2

  2 e x / y 2 + I n y = 2  

y = e 2 e x / e 2 + 1 = 2

x = -e2 In2

New answer posted

2 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

(x12)2+ (y12)2=1

here AB = 2 , BC = 2, AC = 2

area = 12*2*2=1

New answer posted

2 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Tangents making angle π4 with y = 3x + 5.

tanπ4=|m31+3m|m=2, 12

So, these tangents are  . So ASB is a focal chord.

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

  ? d y d x + e x ( x 2 2 ) y = ( x 2 2 x ) ( x 2 2 ) e 2 x

Here, I.F.

= e e x ( x 2 2 ) d x

= e ( x 2 2 x ) e x

 Solution of the differential equation is

y . e ( x 2 2 x ) e x = ( x 2 2 x 1 ) e ( x 2 2 x ) e x

 y(0) = 0

c = 1

y ( 2 ) = 1 + 1 = 0

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

|x29|=3

x=±23, ±6

Required area = A

A2=06 (9x23)dx+03 (9+y9y)dy

A=166+32372=8 [26+439]

Note : No option in the question paper is correct.

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