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Question: The focal chord to \({y^2} = 16x\) is tangent to \({\left( {x - 6} \right)^2} + {y^2} = 2\), then th...

The focal chord to y2=16x{y^2} = 16x is tangent to (x6)2+y2=2{\left( {x - 6} \right)^2} + {y^2} = 2, then the possible value of the slope of this chord are
A). (1,1)\left( { - 1,1} \right)
B). (2,2)\left( { - 2,2} \right)
C). (2,12)\left( { - 2,\dfrac{1}{2}} \right)
D). (2,12)\left( {2, - \dfrac{1}{2}} \right)

Explanation

Solution

Hint: Here, we will proceed by comparing the given equations with general equations of parabola and circle.

The given equation of the parabola is y2=16x (1){y^2} = 16x{\text{ }} \to {\text{(1)}}
The general form of an upward parabola having focus F(a,0){\text{F}}\left( {a,0} \right) is y2=4ax (2){y^2} = 4ax{\text{ }} \to {\text{(2)}}
On comparing above equations (1) and (2), we get
4a=16a=44a = 16 \Rightarrow a = 4
Hence, focus of the given parabola is F(4,0){\text{F}}\left( {4,0} \right)
It is also given that the focal chord of the given parabola is tangent to the circle (x6)2+y2=2(x6)2+(y0)2=(2)2 (3){\left( {x - 6} \right)^2} + {y^2} = 2 \Rightarrow {\left( {x - 6} \right)^2} + {\left( {y - 0} \right)^2} = {\left( {\sqrt 2 } \right)^2}{\text{ }} \to {\text{(3)}}.
Since, the general equation of a circle with centre coordinate as C(x1,y1){\text{C}}\left( {{x_1},{y_1}} \right) and radius as rr is given by (xx1)2+(yy1)2=r2 (4){\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2}{\text{ }} \to {\text{(4)}}
On comparing equations (3) and (4), we get
The centre coordinate of the given circle is C(6,0){\text{C}}\left( {6,0} \right) and radius is 2\sqrt 2 .
Since we know that distance between two points A(a,b){\text{A}}\left( {a,b} \right) and B(c,d){\text{B}}\left( {c,d} \right) is given by distance formula which is d=(ca)2+(db)2d = \sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} .
From the figure,
Distance between two points F(4,0){\text{F}}\left( {4,0} \right) and C(6,0){\text{C}}\left( {6,0} \right) is FC=(64)2+(00)2=(2)2=2{\text{FC}} = \sqrt {{{\left( {6 - 4} \right)}^2} + {{\left( {0 - 0} \right)}^2}} = \sqrt {{{\left( 2 \right)}^2}} = 2
In the figure, we can see that FCT\vartriangle {\text{FCT}} is a right angled triangle with FC as hypotenuse.
Using Pythagoras theorem, we can write
(FC)2=(FT)2+(TC)2(2)2=(FT)2+(r)2(FT)2=4r2=4(2)2=42=2 FT=2  {\left( {{\text{FC}}} \right)^2} = {\left( {{\text{FT}}} \right)^2} + {\left( {{\text{TC}}} \right)^2} \Rightarrow {\left( 2 \right)^2} = {\left( {{\text{FT}}} \right)^2} + {\left( r \right)^2} \Rightarrow {\left( {{\text{FT}}} \right)^2} = 4 - {r^2} = 4 - {\left( {\sqrt 2 } \right)^2} = 4 - 2 = 2 \\\ \Rightarrow {\text{FT}} = \sqrt 2 \\\
 tanθ=PerpendicularBase=TCFT=r2=22=1=tan45 θ=45  \therefore {\text{ }}\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{{\text{TC}}}}{{{\text{FT}}}} = \dfrac{r}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = 1 = \tan {45^ \circ } \\\ \Rightarrow \theta = {45^ \circ } \\\
Since, line FC is a horizontal line.
Also we know that the slope of a line making an angle θ\theta with the horizontal is given by m=tanθm = \tan \theta
Therefore, the slope of the focal chord is m=tanθ=tan45=1m = \tan \theta = \tan {45^ \circ } = 1
or m=tan(180θ)=tan(18045)=tan135=1m = \tan \left( {{{180}^ \circ } - \theta } \right) = \tan \left( {{{180}^ \circ } - {{45}^ \circ }} \right) = \tan {135^ \circ } = - 1
Hence the slope of the required focal chord is (1,1)\left( { - 1,1} \right) which means option A is correct.

Note- In this particular problem line FC is a horizontal line because the y-coordinates of both the points (F and C) are equal. Here, if a line is making an angle θ\theta with respect to the positive x-axis then the angle made by that line with the negative x-axis will be (180θ)\left( {{{180}^ \circ } - \theta } \right). Slope can be measured either with respect to positive x-axis or with respect to negative x-axis.