Question:
<p>A tangent to the hyperbola x<sup>2</sup> - 2y<sup>2</sup> = 4 meets x-axis at P and y-axis at Q. Line PR and QR are drawn such that OPRQ is a rectangle (where O is the origin). The locus of is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{4}{x^{2}}-\frac{2}{y^{2}}\)</span> = 1</p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{x^{2}}-\frac{4}{y^{2}}\)</span> = 1</p>
<p style="display:inline"><span class="math-tex">\(\frac{4}{x^{2}}+\frac{2}{y^{2}}\)</span> = 1</p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{x^{2}}+\frac{4}{y^{2}}\)</span> = 1</p>
Step-by-Step Solution
Key Concept: Identify the intercepts P and Q using the parametric form of the tangent and eliminate the parameter θ using the identity sec²θ - tan²θ = 1 to determine the locus of vertex R.
<p>Equation of tangent to hyperbola <span class="math-tex">$\frac{x^{2}}{4}-\frac{4^{2}}{2}$</span> = 1<br />
is <span class="math-tex">$\frac{\mathrm{x} \sec \theta}{2}-\frac{\mathrm{y} \tan \theta}{\sqrt{2}}$</span> = 1 at any parametric point <span class="math-tex">$\theta$</span><br />
P is (<span class="math-tex">$\frac{2}{\sec \theta}$</span>, 0)<br />
Q is (0, <span class="math-tex">$\frac{-\sqrt{2}}{\tan \theta}$</span>)<br />
R will be x coordinate of r is taken and y-coordinate Q is taken<br />
R<span class="math-tex">$\left(\frac{2}{\sec \theta}, \frac{-\sqrt{2}}{\tan \theta}\right)$</span> h = <span class="math-tex">$\frac{2}{\sec \theta}$</span> K = <span class="math-tex">$\frac{-\sqrt{2}}{\tan \theta}$</span><br />
<span class="math-tex">$\frac{4}{\mathrm{~h}^{2}}=\frac{2}{\mathrm{k}^{2}}$</span> = 1 <span class="math-tex">$\rightarrow$</span> <span class="math-tex">$\frac{4}{x^{2}}-\frac{2}{y^{2}}$</span> = 1</p>
Correct Answer: A