Question

Answer the following:

For the hyperbola `x^2/100−y^2/25` = 1, prove that SA. S'A = 25, where S and S' are the foci and A is the vertex

Answer 1

Given equation of the hyperbola is

`x^2/100 - y^2/25` = 1

Comparing this equation with

`x^2/"a"^2 - y^2/"b"^2` = 1, we get

a² = 100 and b² = 25

∴ a = 10 and b = 5

∴  Co-ordinates of vertex is A(a, 0), i.e., A(10, 0)

Eccentricity e = `sqrt("a"^2 + "b"^2)/"a"`

= `sqrt(100 + 25)/10`

= `sqrt(125)/10`

= `(5sqrt(5))/10`

= `sqrt(5)/2`

Co-ordinates of the foci are S(ae, 0) and S'(–ae, 0)

i.e., `"S"(10(sqrt(5)/2),0)` and `"S'"(-10(sqrt(5)/2),0)`

i.e., `"S"(5sqrt(5), 0)`, and `"S'"(-5sqrt(5), 0)`

Since S, A and S' lie on the X-axis,

SA = `|5sqrt(5) - 10|`

and S'A = `|-5sqrt(5) - 10|`

= `|-5sqrt(5) + 10|`

= `|5sqrt(5) + 10|`

∴ SA. S'A = `|5sqrt(5) - 10| |5sqrt(5) + 10|`

= `|5sqrt(5)^2 - (10)^2|`

= |125 – 100|

= |25|

∴ SA. S'A = 25