<p>Straight line \(Ax + By + D = 0\) would be tangent to \(xy = c^2\), if:</p>
<p>(a) \(A > 0\), \(B > 0\)</p>
<p>(b) \(A < 0\), \(B < 0\)</p>
<p>(c) \(A > 0\), \(B < 0\)</p>
<p>(d) \(A < 0\), \(B > 0\)</p>
Step-by-Step Solution
Key Concept: For a line to be tangent to the hyperbola xy = c², the system of equations must have exactly one solution. This occurs when the discriminant of the resulting quadratic equals zero, which imposes a constraint on the coefficients A, B, and D.
<p><strong>Step 1:</strong> Write the line equation as $Ax + By + D = 0$, so $y = -\frac{Ax + D}{B}$ (assuming $B \neq 0$).</p><p><strong>Step 2:</strong> Substitute into $xy = c^2$: $x \cdot \left(-\frac{Ax + D}{B}\right) = c^2$, which gives $Ax^2 + Dx + Bc^2 = 0$.</p><p><strong>Step 3:</strong> For tangency, the discriminant must equal zero: $\Delta = D^2 - 4A(Bc^2) = 0$, so $D^2 = 4ABc^2$.</p><p><strong>Step 4:</strong> This gives $D = \pm 2c\sqrt{AB}$. For $D$ to be real, we need $AB > 0$.</p><p><strong>Step 5:</strong> The condition $AB > 0$ means either both $A > 0, B > 0$ (option a) or both $A < 0, B < 0$ (option b). Both cases yield real tangent lines.</p><p><strong>Step 6:</strong> When $A$ and $B$ have opposite signs ($AB < 0$), $D$ becomes imaginary, so no real tangent line exists.</p><p><strong>∴ Answer:</strong> a,b</p>
Correct Answer: a,b