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Question
Show that the line 3x – 4y + 10 = 0 is tangent till the hyperbola x2 – 4y2 = 20. Also find the point of contact
Solution
Given equation of the hyperbola is x2 – 4y2 = 20.
∴ `x^2/20 - y^2/5` = 1
Comparing this equation with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,
a2 = 20 and b2 = 5
Given equation of line is 3x – 4y + 10 = 0
∴ y = `(3x)/4 + 5/2`
Comparing this equation with y = mx + c, we get
m = `3/4` and c = `5/2`
For the line y = mx + c to be a tangent to the hyperbola `x^2/"a"^2 - y^2/"b"^2` = 1, we must have
c2 = a2m2 – b2
c2 = `(5/2)^2 = 25/4`
∴ a2m2 – b2 = `20(3/4)^2 - 5`
= `20(9/16) - 5`
= `45/4-5`
= `25/4`
= c2
∴ The given line is a tangent to the given hyperbola and point of contact
= `((-"a"^2"m")/"c", (-"b"^2)/"c")`
= `((-20(3/4))/((5/2)), (-5)/((5/2)))`
= (– 6, – 2)
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