Advertisements
Advertisements
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)`.
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"`.
APPEARS IN
RELATED QUESTIONS
Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Find the area of the region bounded by the ellipse `x^2/4 + y^2/9 = 1.`
The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Find the area bounded by the curve x2 = 4y and the line x = 4y – 2
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Find the area of the smaller region bounded by the ellipse `x^2/a^2 + y^2/b^2 = 1` and the line `x/a + y/b = 1`
Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A (4 , 1), B (6, 6) and C (8, 4).
Find the area of the smaller region bounded by the ellipse \[\frac{x^2}{9} + \frac{y^2}{4} = 1\] and the line \[\frac{x}{3} + \frac{y}{2} = 1 .\]
Area of the region bounded by x2 = 16y, y = 1 and y = 4 and the Y-axis, lying in the first quadrant is _______.
Fill in the blank :
Area of the region bounded by x2 = 16y, y = 1, y = 4 and the Y-axis, lying in the first quadrant is _______.
State whether the following is True or False :
The area bounded by the curve y = f(x), X-axis and lines x = a and x = b is `|int_"a"^"b" f(x)*dx|`.
Choose the correct alternative:
Area of the region bounded by y2 = 16x, x = 1 and x = 4 and the X axis, lying in the first quadrant is ______
Choose the correct alternative:
Area of the region bounded by the parabola y2 = 25x and the lines x = 5 is ______
Find the area of the region bounded by the curve y = `sqrt(9 - x^2)`, X-axis and lines x = 0 and x = 3
Find the area of the region bounded by the curve 4y = 7x + 9, the X-axis and the lines x = 2 and x = 8
Find area of the region bounded by the parabola x2 = 4y, the Y-axis lying in the first quadrant and the lines y = 3
The area bounded by y = `27/x^3`, X-axis and the ordinates x = 1, x = 3 is ______
`int_0^log5 (e^xsqrt(e^x - 1))/(e^x + 3)` dx = ______
`int "e"^x ((sqrt(1 - x^2) * sin^-1 x + 1)/sqrt(1 - x^2))`dx = ________.
Area enclosed between the curve y2(4 - x) = x3 and line x = 4 above X-axis is ______.
Area under the curve `y=sqrt(4x+1)` between x = 0 and x = 2 is ______.
The equation of curve through the point (1, 0), if the slope of the tangent to t e curve at any point (x, y) is `(y - 1)/(x^2 + x)`, is
Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x, is:
The area included between the parabolas y2 = 4a(x +a) and y2 = 4b(x – a), b > a > 0, is
Find the area between the two curves (parabolas)
y2 = 7x and x2 = 7y.
The area of the region bounded by the curve y = sin x and the x-axis in [–π, π] is ______.
The area (in sq. units) of the region {(x, y) : y2 ≥ 2x and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is ______.
The area bounded by the curve, y = –x, X-axis, x = 1 and x = 4 is ______.
Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0,y = 2 and y = 4.
