  Question

The equations of the tangents to the hyperbola $$x^2\, -\, 9\, y^2\, =\, 9$$ that are drawn from $$(3, 2)$$ and the area of the triangle that these tangents form with their chord of contact are:

A
y=512x+34;y2=0;8sq.units  B
y=712x+56;y2=0;6 sq.units  C
y=712x+56;x3=0;8 sq.units  D
y=512x+34;x3=0;8 sq. units  Solution

The correct option is D $$\displaystyle y\, =\, \frac{5}{12}x\, +\, \frac{3}{4};\, x\, -\, 3\, =\, 0;\, 8\,$$ sq. unitsConsider any line with equation $$y=mx+c$$. This line meets the hyperbola $${ x }^{ 2 }-{ 9x }^{ 2 }=9$$ at point whose $$x$$ coordinate are given by $${ x }^{ 2 }-9{ \left( mx+c \right) }^{ 2 }=9$$$$\Rightarrow \left( { 9m }^{ 2 }-1 \right) { x }^{ 2 }+18mcx+9\left( 1+{ c }^{ 2 } \right) =0$$   ...(1)If the line touches the curve, the roots of this equation are equal. In this case$${ \left( 18mc \right) }^{ 2 }=36\left( 9{ m }^{ 2 }-1 \right) \left( 1+{ c }^{ 2 } \right) \Rightarrow { 9m }^{ 2 }=1+{ c }^{ 2 }$$   ...(2)But the point $$\left( 3,2 \right)$$ is lies on  line $$y=mx+c\Rightarrow 2=3m+c$$   ...(3)From (2) and (3) $$\displaystyle { 9m }^{ 2 }=1+{ \left( 2-3m \right) }^{ 2 }\Rightarrow m=\frac { 5 }{ 12 }$$ or $$m$$ is infinityIf $$\displaystyle m=\frac { 5 }{ 12 }$$, then $$\displaystyle c=2-3m=\frac { 3 }{ 4 }$$So the equation of one of the the tangent is $$12y=5x+9$$If $$m\rightarrow \infty$$, the tangent is parallel to the $$y$$ axis, and as it passes through the point $$\left( 3,2 \right)$$ its equation is $$x=3$$Now when the roots of equation (1) are equal $$\displaystyle x=\frac { 18mc }{ 2\left( { 9m }^{ 2 }-1 \right) }$$on this put  $$m =\frac { 5 }{ 12 } ,x=-5$$ , the tangent $$12y=5x+9$$ touches the hyperbola at the point $$\displaystyle \left( -5,\frac { -4 }{ 5 } \right)$$.By observation , the other tangent touches the hyperbola at the vertex $$\left( 3,0 \right)$$The triangle formed by the two angent $$PA$$ and $$PB$$ and chord of contact$$AB$$ has vertices $$\displaystyle \left( 3,2 \right) ,\left( -5,\frac { -4 }{ 3 } \right) ,\left( 3,0 \right)$$The area $$\triangle$$ of triangle $$APB$$ is given by $$\displaystyle 2\triangle ==\begin{vmatrix} 1\quad & 1\quad & 1 \\ 3 & -5 & 3 \\ 2 & \frac { -4 }{ 3 } & 0 \end{vmatrix}\\ \Rightarrow \triangle =8$$Maths

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