<p>Tangents are drawn to the hyperbola \(4x^2 - y^2 = 36\) at the points \(P\) and \(Q\). If these tangents intersect at the point \(T(0, 3)\) then the area (in sq. units) of \(\Delta PTQ\) is</p>
Step-by-Step Solution
Key Concept: For a hyperbola, if tangents from an external point T meet the curve at P and Q, the chord PQ (called chord of contact) can be found using the formula Tx₀x - Ty₀y = a²b². The area of triangle PTQ equals half the product of the chord PQ and the perpendicular distance from T to PQ.
<p><strong>Step 1:</strong> Rewrite hyperbola in standard form: $\frac{x^2}{9} - \frac{y^2}{36} = 1$, so $a^2 = 9$, $b^2 = 36$.</p><p><strong>Step 2:</strong> For point $T(0,3)$ outside the hyperbola, use chord of contact formula: $\frac{x \cdot 0}{9} - \frac{y \cdot 3}{36} = 1$, which gives $-\frac{3y}{36} = 1$, so $y = -12$.</p><p><strong>Step 3:</strong> The chord PQ lies on line $y = -12$. Find points P and Q by substituting into hyperbola: $\frac{x^2}{9} - \frac{144}{36} = 1 \Rightarrow \frac{x^2}{9} = 5 \Rightarrow x = \pm 3\sqrt{5}$.</p><p><strong>Step 4:</strong> So $P(3\sqrt{5}, -12)$ and $Q(-3\sqrt{5}, -12)$ (or vice versa). Length $PQ = 6\sqrt{5}$.</p><p><strong>Step 5:</strong> Distance from $T(0,3)$ to line $y = -12$ is $|3-(-12)| = 15$.</p><p><strong>Step 6:</strong> Area of $\triangle PTQ = \frac{1}{2} \times PQ \times h = \frac{1}{2} \times 6\sqrt{5} \times 15 = 45\sqrt{5}$ sq. units.</p><p>∴ Answer: D</p>
Correct Answer: D