Tangents are drawn to the hyperbola 4x^2 - y^ | vedclass

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(D) The equation of the hyperbola is $4x^2 - y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{36} = 1$.The line $PQ$ is the chord of cont

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$45\sqrt{5}$

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(D) The equation of the hyperbola is $4x^2 - y^2 = 36$,which can be written as $\frac{x^2}{9} - \frac{y^2}{36} = 1$.The line $PQ$ is the chord of contact of the tangents drawn from $T(0, 3)$.The equation of the chord of contact from $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.Substituting $(x_1, y_1) = (0, 3)$,$a^2 = 9$,and $b^2 = 36$,we get:$\frac{x(0)}{9} - \frac{y(3)}{36} = 1$$\Rightarrow -\frac{y}{12} = 1$$\Rightarrow y = -12$.To find the coordinates of $P$ and $Q$,substitute $y = -12$ into the hyperbola equation:$4x^2 - (-12)^2 = 36$$4x^2 - 144 = 36$$4x^2 = 180$$x^2 = 45$$x = \pm 3\sqrt{5}$.Thus,the points are $P(3\sqrt{5}, -12)$ and $Q(-3\sqrt{5}, -12)$.The length of the base $PQ = |3\sqrt{5} - (-3\sqrt{5})| = 6\sqrt{5}$.The height of the triangle $\Delta PTQ$ is the vertical distance from $T(0, 3)$ to the line $y = -12$,which is $h = |3 - (-12)| = 15$.Area of $\Delta PTQ = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 6\sqrt{5} 	imes 15 = 45\sqrt{5}$ sq. units.

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:

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