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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
Solution
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
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