Q70 / 110 · VITEEE 2023 · Mathematics
The area of the triangle formed by the tangent and normal at the point \((1, \sqrt{3})\) on the circle \(\mathrm{x}^2\) \(+y^2=4\) and the \(x\)-axis is
- A3 sq. units
- Correct answerB\(2 \sqrt{3}\) sq. units
- C\(3 \sqrt{2}\) sq. units
- D4 sq. units
Solution
The tangent on \(x^2+y^2=4\) at \((1, \sqrt{3})\) is \([x+\sqrt{3} y=4]\) and equation of normal at \((1, \sqrt{3})\) is \(\mathrm{y}=\mathrm{x} \sqrt{3}\)
\(\begin{aligned} & \text { Required area }=\int_0^1 x \sqrt{3} \mathrm{dx}+\int_1^4 \frac{4-x}{\sqrt{3}} \mathrm{dx} \\ & =\sqrt{3}\left[\frac{\mathrm{x}^2}{2}\right]_0^1+\frac{1}{\sqrt{3}}\left[4 \mathrm{x}-\frac{\mathrm{x}^2}{2}\right]_1^4 \\ & =\sqrt{3} \times \frac{1}{2}+\frac{1}{\sqrt{3}}\left[4(4-1)-\frac{1}{2}(16-1)\right] \\ & =\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}\left[12-\frac{15}{2}\right]=\frac{\sqrt{3}}{2}+\frac{9}{2 \sqrt{3}} \\ & =\frac{\sqrt{3}}{2}+\frac{3 \sqrt{3}}{2}=2 \sqrt{3} \text { sq. units } \end{aligned}\)
\(\begin{aligned} & \text { Required area }=\int_0^1 x \sqrt{3} \mathrm{dx}+\int_1^4 \frac{4-x}{\sqrt{3}} \mathrm{dx} \\ & =\sqrt{3}\left[\frac{\mathrm{x}^2}{2}\right]_0^1+\frac{1}{\sqrt{3}}\left[4 \mathrm{x}-\frac{\mathrm{x}^2}{2}\right]_1^4 \\ & =\sqrt{3} \times \frac{1}{2}+\frac{1}{\sqrt{3}}\left[4(4-1)-\frac{1}{2}(16-1)\right] \\ & =\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}\left[12-\frac{15}{2}\right]=\frac{\sqrt{3}}{2}+\frac{9}{2 \sqrt{3}} \\ & =\frac{\sqrt{3}}{2}+\frac{3 \sqrt{3}}{2}=2 \sqrt{3} \text { sq. units } \end{aligned}\)
Hence, the correct answer is option: 'B'.