Question

using integration find the area of triangle formed by the positive x axis and tangent and normal to the circle x^2+y^2=4 at (1,_/3)

PLZ ANSWER FAST

Answer 1

equation of circle is given,
e.g., x² + y² = 4
we know, equation of tangent of circle x² + y² = C at (x₁,y₁) is given by xx₁ + yy₁ = C
so, equation of tangent through (1,√3) is
x + √3y = 4 -------(1)
=> y = 4/√3 - x/√3 , it cuts the axis at (4,0)

now, equation of normal to the circle is
(y - √3)= slope of normal (x - 1)

[ we know, slope of normal × slope of tangent = -1 so, slope of normal = -1/(-1/√3) = √3 {slope of tangent is -1/√3 as shown equation (1) ]

now, equation of normal is
(y - √3) = √3(y - 1)
=> y - √3 = √3x - √3
=> y = √3x -----(2)

for clearance , you should see attachment,
area formed by tangent, normal and x axis is
$\bold{A = \int\limits^1_0{\sqrt{3}x}\,dx+\int\limits^4_1({\frac{-x}{\sqrt{3}}+\frac{4}{\sqrt{3}})}\,dx}$

$\bold{=\sqrt{3}[\frac{x^2}{2}]^1_0+\frac{1}{\sqrt{3}}[\frac{ - x^2}{2}+4x]^4_1}$

$\bold{=\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}(-8+16 - \frac{ - 1}{2} - 4)}$

$\bold{=\frac{ \sqrt{3}}{2}+\frac{1}{\sqrt{3}}\times\frac{9}{2}}$
$\bold{=2\sqrt{3}}$

hence, answer is 2√3 sq unit

Answer 2

equation of circle is given,
e.g., x² + y² = 4
we know, equation of tangent of circle x² + y² = C at (x₁,y₁) is given by xx₁ + yy₁ = C
so, equation of tangent through (1,√3) is
x + √3y = 4 -------(1)
=> y = 4/√3 - x/√3 , it cuts the axis at (4,0)

now, equation of normal to the circle is
(y - √3)= slope of normal (x - 1)

[ we know, slope of normal × slope of tangent = -1 so, slope of normal = -1/(-1/√3) = √3 {slope of tangent is -1/√3 as shown equation (1) ]

now, equation of normal is
(y - √3) = √3(y - 1)
=> y - √3 = √3x - √3
=> y = √3x -----(2)

for clearance , you should see attachment,
area formed by tangent, normal and x axis is
$\bold{A = \int\limits^1_0{\sqrt{3}x}\,dx+\int\limits^4_1({\frac{-x}{\sqrt{3}}+\frac{4}{\sqrt{3}})}\,dx}$

$\bold{=\sqrt{3}[\frac{x^2}{2}]^1_0+\frac{1}{\sqrt{3}}[\frac{ - x^2}{2}+4x]^4_1}$

$\bold{=\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}(-8+16 - \frac{ - 1}{2} - 4)}$

$\bold{=\frac{ \sqrt{3}}{2}+\frac{1}{\sqrt{3}}\times\frac{9}{2}}$
$\bold{=2\sqrt{3}}$

hence, answer is 2√3 sq unit