Question

The area of triangle formed by the tangents from point $$(3, 2)$$ to hyperbola $$x^2-9y^2=9$$ and the chord of contact w.r.t. point $$(3, 2)$$ is :

A
8
B
4
C
6
D
16

Solution

The correct option is D $$8$$Hyperbola is $$x^2-9y^2=9$$ or $$\dfrac {x^2}{9}-\dfrac {y^2}{1}=1$$Equation of tangent is $$y=mx\pm \sqrt {a^2m^2-b^2}$$               ..... (i)It passes through $$(3, 2)$$$$\Rightarrow 2=3m\pm \sqrt {9m^2-1}$$or $$4+9m^2-12m=9m^2-1$$$$\Rightarrow m_1=\dfrac {5}{12}$$ and $$m_2=\infty$$Equation of tangent (i) for $$m_1=\dfrac {5}{12}$$$$y=\dfrac {5}{12}x\pm \sqrt {9(\dfrac {5}{12})^2-1}$$or $$y=\dfrac {5}{12}x\pm \dfrac {3}{4}$$On taking (-)ve sign, point $$P(3, 2)$$ does not satisfy the equation of tangent therefore rejecting (-)ve sign. Hence equation of tangent is $$y=\dfrac {5x}{12}+\dfrac {3}{4}$$                   .....(ii)Now, equation of tangent (i) for $$m_2=\infty$$ is $$x\pm 3=0$$rejecting $$(+)$$ sign (since taking $$(+)$$ sign point $$P(3, 2)$$ does not satisfy this equation.)Hence, equation of tangent is $$x-3=0$$                   ....(iii)Now equation of chord of contact w.r.t. point $$P(3, 2)$$ is $$T=0$$or $$3x-18y=9$$or $$x-6y=3$$                                             .....(iv)Solving (ii) and (iv); $$x=-5, y=-\dfrac {4}{3}$$Solving (iii) and (iv); $$x=3, y=0$$Now vertices of triangle are $$\displaystyle(3, 2), (3, 0), (-5, -\dfrac{4}{3})$$$$\therefore Area=\dfrac {1}{2}\begin{vmatrix}3 & 2 & 1\\ 3 & 0 & 1\\ -5 & -\dfrac {4}{3} & 1\end{vmatrix}$$$$=\dfrac {1}{2}|4-16-4|$$$$=8 sq. units$$Maths

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