Question

Using integration find the area of the triangle formed by positive x-axis and tangent and normal to the circle $x^2+y^2=4$ at $(1,\sqrt{3})$.

Answer 1

Given circle is $x^2+y^2=4$

$\Rightarrow 2x+2y\frac{dy}{dx}=0$ [ By differentiating]$\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$

Now, slope of tangent at $(1,\sqrt{3})={\left[\frac{dy}{dx}\right]}_{(1,\sqrt{3})}=-\frac{1}{\sqrt{3}}$

$\therefore$ Slope of normal at $\left(1,\sqrt{3}\right)=\sqrt{3}$

Therefore, equation of tangent is

$\frac{y-\sqrt{3}}{x-1}=-\frac{1}{\sqrt{3}}$

$\Rightarrow x+\sqrt{3}y=4$ ...(i)

Equation of normal is

$\frac{y-\sqrt{3}}{x-1}=\sqrt{3}$

$\Rightarrow y-\sqrt{3}x=0$ ...(ii)

To draw the graph of the triangle formed by the x-axis, line (i) and line (ii), we find the intersecting points of these three lines which give vertices of the required triangle. Let O, A. B be the intersecting points of these lines.

The coordinate of O, A, B are $(0,0),(1,\sqrt{3})$ and $(4,0)$ respectively.

$\therefore$ Required area $=$ area of triangle $OAB=$ area of region $OAC+$ area of region $CAB$

$={\int }_0^{1}ydx+{\int }_1^{4}ydx$ [Where in 1st integrand, $y=\sqrt{3}x$ and in 2nd, $y=\frac{4-x}{\sqrt{3}}$]

$={\int }_0^1\sqrt{3x}dx+{\int }_1^4\frac{4-x}{\sqrt{3}}dx=\sqrt{3}{\left[\frac{x^2}{2}\right]}_0^1-\frac{1}{\sqrt{3}}{\left[\frac{(4-x{)}^2}{2}\right]}_1^4$

$=\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{3}}[0-\frac{9}{2}]$

$=\frac{\sqrt{3}}{2}+\frac{9}{2\sqrt{3}}=\frac{12}{2\sqrt{3}}=2\sqrt{3}$ sq. units.