Tangents are drawn to the hyperbola x^2 - 9y^ | vedclass

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(D) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{1} = 1$,where $a^2 = 9$ and $b^2 = 1$.The equation of a tangent with slope $m$ is $y

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$8$

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(D) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{1} = 1$,where $a^2 = 9$ and $b^2 = 1$.The equation of a tangent with slope $m$ is $y = mx \pm \sqrt{9m^2 - 1}$. Since it passes through $(3, 2)$,we have $2 = 3m \pm \sqrt{9m^2 - 1}$.$(2 - 3m)^2 = 9m^2 - 1$ $\Rightarrow 4 - 12m + 9m^2 = 9m^2 - 1$ $\Rightarrow 12m = 5$ $\Rightarrow m = \frac{5}{12}$.The other tangent is the vertical line $x = 3$ (since the point $(3, 2)$ lies on the line $x = 3$).The equation of the first tangent is $y - 2 = \frac{5}{12}(x - 3)$ $\Rightarrow 12y - 24 = 5x - 15$ $\Rightarrow 5x - 12y + 9 = 0$.The chord of contact for the point $(x_1, y_1) = (3, 2)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$ $\Rightarrow \frac{3x}{9} - \frac{2y}{1} = 1$ $\Rightarrow x - 6y = 3$.The vertices of the triangle are the intersection of the two tangents $(3, 2)$,the intersection of the first tangent and the chord of contact $(3, 1)$,and the intersection of the second tangent $(x=3)$ and the chord of contact $(3, 0)$. Wait,let us re-evaluate the intersection points.Intersection of $x=3$ and $x-6y=3$ is $(3, 0)$.Intersection of $5x-12y+9=0$ and $x-6y=3$ (i.e.,$x=6y+3$): $5(6y+3) - 12y + 9 = 0$ $\Rightarrow 30y + 15 - 12y + 9 = 0$ $\Rightarrow 18y = -24$ $\Rightarrow y = -4/3$. Then $x = 6(-4/3) + 3 = -8 + 3 = -5$.The vertices are $(3, 2)$,$(3, 0)$,and $(-5, -4/3)$.Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{1}{2} |3(0 - (-4/3)) + 3(-4/3 - 2) + (-5)(2 - 0)| = \frac{1}{2} |3(4/3) + 3(-10/3) - 10| = \frac{1}{2} |4 - 10 - 10| = \frac{1}{2} |-16| = 8$ sq units.

Tangents are drawn to the hyperbola $x^2 - 9y^2 = 9$ from the point $(3, 2)$. The area of the triangle formed by the tangents and the chord of contact is . . . . . . sq units.

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