Question

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

• A. $45\sqrt{5}$
• B. $54\sqrt{3}$
• C. $60\sqrt{3}$
• D. $36\sqrt{5}$

Answer 1

The correct option is A $45\sqrt{5}$

$4x^2-y^2=36$ $T(0,3)$

$8x-2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{4x}{y}=m$

$y=mx+c\Rightarrow y=\frac{4x^2}{y}+c$

${}^y{|}_{(0,3)}=\frac{4x^2}{y}+c\Rightarrow c=3$

$\Rightarrow$ equation of tangent $\Rightarrow y=\frac{4x^3}{y}+3$ ie $4x^2=y^2-3y$

The pt where the tangents from pt T(0,3)

touch the hyperbola satisfy the equations :

$4x^2-y^2=36$ and $4x^2=y^2-3y$

$y^2+36=4x^2=y^2-3y$

$y^2+36=y^2-3y\Rightarrow y=-12$

$4x^2=y^2-3y\Rightarrow x\pm 3\sqrt{5}$

P & Q $=(-3\sqrt{5},-12)$

$(3\sqrt{5},-12)$

base of $△PTQ=6\sqrt{5}$

height $=3+12=15$

area of $△PTQ=\frac{1}{2}\times 6\sqrt{5}\times 15=45\sqrt{5}uni{t}^2$