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Question
Using integration find the area of the triangle formed by negative x-axis and tangent and normal to the circle `"x"^2 + "y"^2 = 9 "at" (-1,2sqrt2)`.
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Solution
The equation of circle is `"x"^2 + "y"^2 = 9`
∴ `2"x" + 2"y"(d"y")/(d"x") = 0`
⇒ `(d"y")/(d"x") = -("x")/("y")`
Slope of tangent at `(-1,2sqrt(2)) and "m"_"T" = (1)/(2sqrt(2)`
∴ eq. of tangent : `"y" - 2sqrt(2) = (1)/(2sqrt(2))("x"+ 1)`
⇒ `"x" - 2sqrt(2) "y" + 9 = 0`
Clearly it cuts x axis at `(-9,0)`
Also eq. of normal : `"y"-2sqrt(2) = -2sqrt(2)("x"+1)`
⇒ `2sqrt(2"x")+"y" = 0`
As tangent and normal both meet at the point `(-1,2sqrt(2))`.
So, ar(OPB) = `int_-9^-1 ("x"+9)/(2sqrt(2))d"x" + int_-1^0 -2sqrt(2)"x"d"x"`
⇒ = `(1)/(2sqrt(2))["x"^2/(2) + 9"x"]_-9^-1 - sqrt2["x"^2/2]_-1^0`
⇒ = `(1)/(2sqrt2)[(1/2 - 9)-(81/2 - 81)]-sqrt(2)[0-1]`
∴ ar(OPB) = `9sqrt2 "sq.units"`.
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