Question:
<p>The tangent and the normal lines at the point <span class="math-tex">\((\sqrt{3}, 1)\)</span> to the circle x<sup>2</sup> + y<sup>2</sup> = 4 and the X-axis form a triangle. The area of this triangle (in square units) is</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\sqrt{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{4}{\sqrt{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{\sqrt{3}}\)</span></p>
Step-by-Step Solution
Key Concept: The area of the triangle is determined by the x-intercepts of the tangent and normal lines which form the base, and the y-coordinate of the point of tangency which acts as the height.
<p>Let T = 0 and N = 0 represents the tangent and normal lines at the point p (<span class="math-tex">$\sqrt{3}$</span>, 1) to the circle x<sup>2</sup> + y<sup>2</sup> = 4<br />
<img alt="" data-imgur-src="JHrVfR3.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/JHrVfR3.png" style="width: 250px; height: 195px;" /><br />
So, equation of tangent (T = 0) is<br />
<span class="math-tex">$\sqrt{3}$</span> x + y = 4 .....(i)<br />
For point A, put y = 0, we get<br />
x = <span class="math-tex">$\frac{4}{\sqrt{3}}$</span><br />
<span class="math-tex">$\because$</span> Area of required <span class="math-tex">$\triangle$</span>OPA = <span class="math-tex">$\frac{1}{2}$</span> (OA)(PM)<br />
= <span class="math-tex">$\frac{1}{2} \times \frac{4}{\sqrt{3}} \times 1$</span> [<span class="math-tex">$\because$</span> PM = y-coordinate of P]<br />
= <span class="math-tex">$\frac{2}{\sqrt{3}}$</span>sq unit</p>
Correct Answer: D