Conic Sections

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New answer posted

6 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

4x – 3y + k2 = 0

2 r = 4 0 5 = 8 r = 4

( x 1 ) 2 + ( y + 2 ) 2 = 1 6

New question posted

6 months ago

0 Follower 4 Views

New answer posted

6 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

Equation of tangent to the circle at (2, 4) is

(4 + A) x + (8 + B) y + 2A + 2B + 2C = 0……. (i)

Also equation of tangent to parabola y = x2 at (2, 4) is

4x – y = 4 ………. (ii)

Comparing (i) and (ii) we get, A + C = 16

New answer posted

6 months ago

0 Follower 82 Views

P
Payal Gupta

Contributor-Level 10

S (4, 4) and V (3, 2)

 point of intersection of directrix with axis of parabola is A (2, 0)

Image of A (2, 0) with respect to line

x+2y=6isB (x2, y2)

x221=y2022 (2+06)5

B (185, 165)

Point B is point of intersection of directrix with axes of parabola P2.

x+2y=λ

B (185, 165) lies on the line x + 2y = λ  λ = 1 8 5 + 3 2 5 = 1 0

New answer posted

6 months ago

0 Follower 6 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

6 months ago

0 Follower 5 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

6 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Required area

42 (4yy22)dy

(4yy22y36)42=18squnits

New answer posted

6 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

 x2a2+y24=1

Δ=12*a (1+cosθ).4sinθ

ΔmaxΔ=63a=4

e=32

 

New answer posted

6 months ago

0 Follower 4 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

6 months ago

0 Follower 3 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

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