Tangents are drawn to the hyperbola 4x2 - y2 = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of Δ PTQ is :

Question

Tangents are drawn to the hyperbola 4x2 - y2 = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of Δ PTQ is : (A) 45 √5 (B) 54√3 (C) 60√3 (D) 36√5

Tardigrade Admin · Accepted Answer

Clearly P Q is a chord of contact, i.e., equation of P Q is T ≡ 0 ⇒ y=-12 Solving with the curve, 4 x2-y2=36 ⇒ x=± 3 √5, y=-12 i.e., P(3 √5,-12) ; Q(-3 √5,-12) ; T(0,3) Area of Δ P Q T is Δ=(1/2) × 6 √5 × 15 =45 √5 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/f77718f5b5d917805d135009f9cb212a-.png" />

Q. Tangents are drawn to the hyperbola $4 x^{2} - y^{2} = 36$ at the points $P$ and $Q$ . If these tangents intersect at the point $T (0, 3)$ then the area (in sq. units) of $Δ PTQ$ is :

$455$

$543$

$603$

$365$

Clearly $PQ$ is a chord of contact,
i.e., equation of $PQ$ is $T \equiv 0$
$\Rightarrow y = - 12$
Solving with the curve, $4 x^{2} - y^{2} = 36$
$\Rightarrow x = \pm 35, y = - 12$
i.e., $P (35, - 12); Q (- 35, - 12); T (0, 3)$
Area of $Δ PQT$ is
$Δ = \frac{1}{2} \times 65 \times 15$
$= 455$

Q. Tangents are drawn to the hyperbola 4x2−y2=36 at the points P and Q. If these tangents intersect at the point T(0,3) then the area (in sq. units) of ΔPTQ is :

455​

543​

603​

365​