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Question
Show that the line x – y = 5 is a tangent to the ellipse 9x2 + 16y2 = 144. Find the point of contact
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Solution
Given equation of the ellipse is 9x2 + 16y2 = 144
∴ `x^2/16 + y^2/9` = 1
Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1, we get
a2 = 16 and b2 = 9
Given equation of line is x – y = 5, i.e., y = x – 5
Comparing this equation with y = mx + c, we get
m = 1 and c = – 5
For the line y = mx + c to be a tangent to the ellipse `x^2/"a"^2 + y^2/"b"^2` = 1, we must have
c2 = a2m2 + b2
c2 = (–5)2 = 25
a2m2 + b2 = 16(1)2 + 9 = 16 + 9 = 25 = c2
∴ The given line is a tangent to the given ellipse and point of contact
= `((-"a"^2"m")/"c", "b"^2/"c")`
= `(((-16)(1))/-5, 9/-5)`
= `(16/5, (-9)/5)`.
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