Question

The hyperbola `x^2/a^2 - y^2/b^2` = 1 passes through the point of intersection of the lines, 7x + 13y – 87 = 0 and 5x – 8y + 7 = 0, the latus rectum is `32sqrt(2)/5`. The value of `(asqrt(2) + b)` will be ______.

Options

• 7.00

• 8.00

• 9.00

• 10.00

Answer 1

The hyperbola `x^2/a^2 - y^2/b^2` = 1 passes through the point of intersection of the lines, 7x + 13y – 87 = 0 and 5x – 8y + 7 = 0, the latus rectum is `32sqrt(2)/5`. The value of `(asqrt(2) + b)` will be 9.00.

Explanation:

`x^2/a^2 - y^2/b^2` = 1 is given hyperbola point of intersection

35x + 65y – 435 = 0
35x – 56y +   49 = 0
   –   +          –                   
        121y            = 484

y = 4

∴ 5x = 8y – 7

∴ 5x = 32 – 7 = 25

x = 5

∴ (5, 4) is the point of intersection through which hyperbola passes `25/a^2 - 16/b^2` = 1

⇒ `25/a^2 = 1 + 16/b^2`  ...(i)

and `(2b^2)/a = (32sqrt(2))/5`

⇒ b² = `(16sqrt(2)a)/5`

From (i) and (ii),

`25/a^2 = 1 + 16/b^2`

⇒ `25/a^2 = 1 + 5/(asqrt(2))`

⇒ `(5/a)^2 - 5/(asqrt(2)) - 1` = 0

⇒ `(5/a - 1/(2sqrt(2)))^2 = 1 + 1/8 = 9/8`

`5/a - 1/(2sqrt(2)) = 3/(2sqrt(2))`

`5/a = 1/(2sqrt(2)) + 3/(2sqrt(2))`

`5/a = 4/(2sqrt(2)) = sqrt(2)`

`5/a = sqrt(2)`

a = `5/sqrt(2)`

∴ b² = `(16sqrt(2))/5.a`

∴ b² = `(16sqrt(2))/5. 5/sqrt(2)`

∴ b² = 16

b = 4

Now, `(asqrt(2) + b) = 5/sqrt(2) xx sqrt(2) + 4` = 9