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 PTQ is :

Engineering

Mathematics

Tangent to Hyperbola

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 $∆$ PTQ is :

$45 \sqrt{5}$

$54 \sqrt{3}$

$60 \sqrt{3}$

$36 \sqrt{5}$

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$\frac{x^{2}}{9} - \frac{y^{2}}{36} = 1$

y = mx ± $\sqrt{9 m^{2} - 36}$

Tangent passes through (0, 3)

3 = ± $\sqrt{9 m^{2} - 36}$

9 = 9m² – 36

m = ± $\sqrt{5}$

Equation of PT

$y = \sqrt{5} x + 3 \Rightarrow \frac{- \sqrt{5} x}{3} + \frac{y}{3} = 1$ ….…(1)

Let P be (x₁ , y₁)

$\frac{x x_{1}}{9} - \frac{y y_{1}}{36} = 1$ ….…(2)

Comparing (1) & (2)

$\frac{x_{1}}{9} = \frac{- \sqrt{5}}{3} \Rightarrow x_{1} = - 3 \sqrt{5}$

$\frac{- y_{1}}{36} = \frac{1}{3}$ ⇒ y₁ = –12,

$∴$ P (– 3 $\sqrt{5}$ , – 12)

$∴$ Required area = $\frac{1}{2} \times 15 \times 6 \sqrt{5} = 45 \sqrt{5}$ .