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Question
Show that the line 8y + x = 17 touches the ellipse x2 + 4y2 = 17. Find the point of contact
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Solution
Given equation of the ellipse is x2 + 4y2 = 17.
∴ `x^2/17 + y^2/(17/4)` = 1
Comparing this equation with `x^2/"a"^2 + y^2/"b"^2` = 1 we get
a2 = 17 and b2 = `17/4`
Given equation of line is 8y + x = 17,
i.e., y = `(-1)/8 "x" + 17/8`
Comparing this equation with y = mx + c, we get
m = `(-1)/8` and c = `17/8`
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 = a2 m2 + b2
c2 = `(17/8)^2 = 289/64`
a2m2 + b2 = `17((-1)/8)^2 + 17/4`
= `17/64 + 17/4`
= `289/64`
= c2
∴ The given line touches the given ellipse and point of contact is
`((-"a"^2"m")/"c", "b"^2/"c") = ((-17((-1)/8))/(17/8), (17/4)/(17/8))`
= (1, 2)
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