Question
The focal chord to $${y^2} = 16x$$ is tangent to $${\left( {x - 6} \right)^2} + {y^2} = 2,$$ then the possible values of the slope of this chord, are-
A.
$$\left\{ { - 1,\,1} \right\}$$
B.
$$\left\{ { - 2,\,2} \right\}$$
C.
$$\left\{ { - 2,\, - \frac{1}{2}} \right\}$$
D.
$$\left\{ {2,\, - \frac{1}{2}} \right\}$$
Answer :
$$\left\{ { - 1,\,1} \right\}$$
Solution :
For parabola $${y^2} = 16x,$$ focus $$ \equiv \left( {4,\,0} \right).$$
Let $$m$$ be the slope of focal chord then equation is
$$y = m\left( {x - 4} \right).....(1)$$
But given that above is a tangent to the circle
$${\left( {x - 6} \right)^2} + {y^2} = 2$$
With Centre, $$C\left( {6,\,0} \right),\,r = \sqrt 2 $$
$$\therefore $$ Length of $${ \bot ^{{\text{lar}}}}$$ from (6, 0) to (1) $$=r$$
$$\eqalign{
& \Rightarrow \frac{{6m - 4m}}{{\sqrt {{m^2} + 1} }} = \sqrt 2 \cr
& \Rightarrow 2m = \sqrt {2\left( {{m^2} + 1} \right)} \cr
& \Rightarrow 2{m^2} = {m^2} + 1 \cr
& \Rightarrow {m^2} = 1 \cr
& \Rightarrow m = \pm 1 \cr} $$