<p>The equation of a tangent to the hyperbola \(x^2 - 2y^2 = 18\) which is perpendicular to the line \(x - y = 0\) is</p>
<p>(a) \(x + y = 3\)</p>
<p>(b) \(x + y + 3 = 0\)</p>
<p>(c) \(x + y + 3\sqrt{2} = 0\)</p>
<p>(d) \(x + y + 3\sqrt{2} = 0\)</p>
Step-by-Step Solution
Key Concept: For a hyperbola, use the condition that tangent slope m satisfies c² = a²m² - b², and match this with the perpendicularity constraint (slope = -1 since line is y = x).
<p><strong>Step 1:</strong> Rewrite hyperbola in standard form: $\frac{x^2}{18} - \frac{y^2}{9} = 1$</p><p>Here $a^2 = 18$, $b^2 = 9$</p><p><strong>Step 2:</strong> Line $x - y = 0$ has slope 1. Perpendicular line has slope $m = -1$</p><p><strong>Step 3:</strong> For tangent to hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$, use:</p><p>$$c^2 = a^2m^2 - b^2$$</p><p><strong>Step 4:</strong> Substitute $m = -1$, $a^2 = 18$, $b^2 = 9$:</p><p>$$c^2 = 18(1) - 9 = 9$$</p><p>$$c = \pm 3$$</p><p><strong>Step 5:</strong> Equation of tangent: $y = mx + c = -x \pm 3$</p><p>$$x + y - 3 = 0 \text{ or } x + y + 3 = 0$$</p><p>∴ Answer: C</p>
Correct Answer: C