Conic Sections

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New answer posted

6 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

2x + y = 1

m 1 = 2 1 = 2 a n d m 1 m 2 = 1          

2 . m 2 = 1           

m 2 = 1 2           

y2 = 6x

y2 = 4 (3/2) x

y = m x + a m         

y = 1 2 x + 3 2 1 2        

y = x 2 + 3           

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

x225+y216=1

16 = 25 (1 – e2)

e=35

OF1 = 5  (35)=3

For Hyperbola : e' =53

a = 3

Hyperbola, x29y216=1

New answer posted

6 months ago

0 Follower 14 Views

P
Payal Gupta

Contributor-Level 10

Combined equation of pair of lines OP and OQ is

x2+2y2=2 (x+y)2

x2+4xy=0x (x+4y)=0 {x=0 (lineOP)y=x4 (lineOQ)

tan (90°+θ)=14

cotθ=14tanθ=4

θ=tan14=cot114=π2tan114

POQ=πθ=π2+tan114

New answer posted

6 months ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

Using the standard equations of a hyperbola:

9e2+l and directrix focusae=10

By multiplying both focus and directrix, we get
ae=910 and a2=9
Now e=103
(ae)2=a2+b2

New answer posted

6 months ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

E 1 : x 2 a 2 + y 2 b 2 = 1

  e 1 2 = 1 b 2 a 2 -(i)

Let E 2 : x 2 a 2 + y 2 B 2 = 1  

B > a

  e 2 2 = 1 a 2 B 2 a n d g i v e n B e 2 = b

e 2 = 3 5 2 = ( 5 1 2 ) 2

e = 5 1 2

New answer posted

6 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

L : 2x + y = k.

y = 2 x + k . i s t a n g e n t x 2 3 y 2 3 = 1                

k 2 = 3 ( 2 ) 2 3 = 9 k = 3 a s k > 0              

y = -2x + 3 is also tangent to y2 = 4 ( α 4 ) x  

3 = α / 4 2 -> a = -24

New answer posted

6 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Centre of the smallest circle is A

Centre of the largest circle is B

r 1 = | C P + C B | = 3 2 + 3 and

r 2 = | C P C B | = 3 2 3    

r 1 r 2 = 3 2 + 3 3 2 3 = 3 + 2 2

New answer posted

6 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Let mid point of PQ is R (h, k)

h = α + 4 α 2 + α + 1 2 2 a n d

   

k = 4 α 2 + 1 + 4 α 2 + α + 1 2 2   

Eliminate a from above these two, we get

2 (3x – y)2 + (x – 3y) + 2 = 0

New answer posted

6 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  ( p q ) = p q

 

New answer posted

6 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

A P = 9 + 1 6 + 4 + 1

AP = 3 = AQ

r = 1 + 4 1 = 2        

t a n θ = 3 2  

A r e a o f Δ A P Q A r e a o f Δ B P Q = A R R B = 3 s i n θ 2 c o s θ = 9 4

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