# Answer the following: For the hyperbola x2100−y225 = 1, prove that SA. S'A = 25, where S and S' are the foci and A is the vertex - Mathematics and Statistics

Sum

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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#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 7 Conic Sections
Miscellaneous Exercise 7 | Q 2.16 | Page 178