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) \[\left\{ { - 2,1/2} \right\}\]
D) \[\left\{ {2, - 1/2} \right\}\]

Answer 1

Hint:
Here we will first find the focus point of the parabola. Then we will find the equation of the focal chord passing through the focus of the parabola. We will then find the value of the center coordinates of the circle and the radius. Using the relation of the distance of the tangent to the circle equal to the radius of the circle, we will solve the equation to get the possible values of the slope of this chord.

Complete Step by Step Solution:
Given equation of the parabola is \[{y^2} = 16x\].
Now we will compare this given equation of the parabola with the original form of the equation of parabola to get the value of focus of the parabola. We know that the equation of parabola is \[{y^2} = 4ax\] with the focus of the parabola is \[\left( {a,0} \right)\]. Therefore, we get
Focus point of the given parabola is \[\left( {4,0} \right)\].
Now we will find the equation of the line i.e. focal chord which is passing through the point \[\left( {4,0} \right)\]. Therefore, we get
Equation of the line i.e. focal chord passing through the point \[\left( {4,0} \right)\] is \[\left( {y - 0} \right) = m\left( {x - 4} \right)\]
Equation of the line i.e. focal chord passing through the point \[\left( {4,0} \right)\] is \[mx - y - 4m = 0\]
Now we will find the center coordinates of the circle and the radius of the circle from the given equation of the circle.
Given equation of the circle is \[{\left( {x - 6} \right)^2} + {y^2} = 2\]
Now we will compare the given equation of the circle with the standard equation of the circle. The equation of the circle i.e. \[{\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}\] where, \[{x_0}\] is the \[x\] coordinate of the center of the circle and\[{y_0}\] is the \[y\] coordinate of the center of the circle and \[r\] is the radius of the circle. Therefore, we get
Center of the circle \[\left( {6,0} \right)\] and radius of the circle is \[\sqrt 2 \].
It is given that the focal chord is tangent to the circle which means that the distance of the focal chord from the center of the circle is equal to the radius of the circle. Therefore, we get
\[\left| {\dfrac{{mx - y - 4m}}{{\sqrt {1 + {m^2}} }}} \right| = \sqrt 2 \]
Now we will put the value of \[x = 6\] and \[y = 0\] in the above equation, we get
\[ \Rightarrow \left| {\dfrac{{6m - 0 - 4m}}{{\sqrt {1 + {m^2}} }}} \right| = \sqrt 2 \]
Now we will solve this equation to get the value of \[m\].
\[ \Rightarrow \left| {\dfrac{{6m - 4m}}{{\sqrt {1 + {m^2}} }}} \right| = \sqrt 2 \]
Squaring both side of the equation, we get
\[ \Rightarrow 36{m^2} + 16{m^2} - 48{m^2} = 2 \times \left( {1 + {m^2}} \right)\]
\[ \Rightarrow 52{m^2} - 48{m^2} = 2 + 2{m^2}\]
Adding and subtracting the like terms, we get
\[ \Rightarrow {m^2} = 1\]
Taking square root on the both sides, we get
\[ \Rightarrow m = \pm 1\]
Hence the possible values of the slope of this chord are \[\left\{ { - 1,1} \right\}\].

Therefore, option A is the correct option.

Note:
The chord of a circle is the line segment whose endpoint always lies on the circumference of the circle. We know that the diameter of the circle is the longest chord that passes through the center of the circle. In other words, we can say that the chords passing through the center of the circle are always the longest chord of that circle. Also, the radius of the circle is not known as the chord as one endpoint lies in the center of the circle. We should also remember that when a point lies on a curve or a line then that point satisfies the equation of that curve or line.