Math, asked by Anonymous, 1 month ago

Find the equation of the hyperbola whose asymptotes are x + 2y + 3 = 0, 3x + 4y + 5 = 0 and which passes through the point (1, -1).

Answers

Answered by jaiswalritesh5142
0

Equation of a hyperbola and its asymptotes differ in constant term only.

equation of the hyperbola with asymptotes

x

+

2

y

+

3

=

0

and

3

x

+

4

y

+

5

=

0

can be represented as

(

x

+

2

y

+

3

)

(

3

x

+

4

y

+

5

)

+

λ

=

0

We know that point

(

1

,

1

)

lies on the hyperbola

(

1

+

2

×

(

1

)

+

3

)

(

3

×

1

+

4

×

(

1

)

+

5

)

+

λ

=

0

λ

=

8

Substituting value of

λ

, equation of the hyperbola is

(

x

+

2

y

+

3

)

(

3

x

+

4

y

+

5

)

8

=

0

or

3

x

2

+

8

y

2

+

10

x

y

+

14

x

+

22

y

+

7

=

0

We know that

H

+

C

=

2

A

or

C

=

2

A

H

, where H is Equation of Hyperbola, C is its conjugate and A is the asymptotes

Solving the above, we can get the equation of the conjugate as

3

x

2

+

8

y

2

+

10

x

y

+

14

x

+

22

y

+

23

=

0

Answered by CopyThat
28

Step-by-step explanation:

Equation of the asymptotes are :

  • x + 2y + 3 = 0
  • 3x + 4y + 5 = 0

Equation of the hyperbola can be taken as :

  • (x + 2y + 3)(3x + 4y + 5) + k = 0

The hyperbola passes through P(1,-1) :

  • (1 - 2 + 3)(3 - 4 + 5) + k = 0
  • k = -8

Equation of the hyperbola is (x + 2y + 3)(3x + 4y + 5) - 8 = 0

⇒ 3x² + 6xy + 9x + 4xy + 8y² + 12y + 5x + 10y + 15 - 8 = 0

3x² + 10xy + 8y² + 14x + 22y + 7 = 0

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