मराठी

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 , 2 √ 2 ) .

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प्रश्न

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)`.

बेरीज
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उत्तर

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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2015-2016 (March) All India Set 1 E

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