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Question: The focal chord to \({y^2} = 64x\) is a tangent to \({\left( {x - 4} \right)^2} + {\left( {y - 2} \r...

The focal chord to y2=64x{y^2} = 64x is a tangent to (x−4)2+(y−2)2=4{\left( {x - 4} \right)^2} + {\left( {y - 2} \right)^2} = 4 then the possible values of the slope of this chord is
a.0,−12350, - \dfrac{{12}}{{35}}
b.0,12350,\dfrac{{12}}{{35}}
c.0,−35120, - \dfrac{{35}}{{12}}
d.0,−6350, - \dfrac{6}{{35}}

Explanation

Solution

We are given a equation of a parabola and comparing it with the general equation y2=4ax{y^2} = 4ax where ( a , 0) is the focus ,we get the focus and same way we are given a equation of a circle and comparing it with the general equation (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}where (h,k)\left( {h,k} \right) is the centre and r is its radius ,we get its centre and using the general equation of the tangent to a circle ⇒(y−k)=m(x−h)+21+m2\Rightarrow \left( {y - k} \right) = m\left( {x - h} \right) + 2\sqrt {1 + {m^2}} , where m is the slope. As the tangent is a focal chord it passes through the focus hence it should satisfy the equation and hence the required slope is obtained .

Complete step-by-step answer:
We given the equation of a parabola y2=64x{y^2} = 64x
We know that the general equation of the parabola is y2=4ax{y^2} = 4axwhere ( a , 0) is the focus
Comparing the equations we get
⇒4a=64 ⇒a=644=16  \Rightarrow 4a = 64 \\\ \Rightarrow a = \dfrac{{64}}{4} = 16 \\\
Hence from this we get the focus of the parabola to be (16,0)\left( {16,0} \right)
We are also given an equation of a circle (x−4)2+(y−2)2=4{\left( {x - 4} \right)^2} + {\left( {y - 2} \right)^2} = 4
We know the general equation of a circle is (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}where (h,k)\left( {h,k} \right)is the centre and r is its radius
Comparing the equations we get the centre to be (4,2)\left( {4,2} \right)
We are given that the focal chord is a tangent to the given circle
Now we know that the equation of a tangent to a circle is given by
⇒(y−k)=m(x−h)+21+m2\Rightarrow \left( {y - k} \right) = m\left( {x - h} \right) + 2\sqrt {1 + {m^2}} ………..(1)
Where m is the slope of the tangent
Let's use the centre in (1)
⇒(y−2)=m(x−4)+21+m2\Rightarrow \left( {y - 2} \right) = m\left( {x - 4} \right) + 2\sqrt {1 + {m^2}}
Since we are given that the tangent is a focal chord
It clearly tells that the tangent passes through the focus(16,0)\left( {16,0} \right)
It means that satisfies the above equation
⇒(0−2)=m(16−4)+21+m2 ⇒−2=m(12)+21+m2 ⇒−2−12m=+21+m2 ⇒−2(1+6m)=+21+m2  \Rightarrow \left( {0 - 2} \right) = m\left( {16 - 4} \right) + 2\sqrt {1 + {m^2}} \\\ \Rightarrow - 2 = m\left( {12} \right) + 2\sqrt {1 + {m^2}} \\\ \Rightarrow - 2 - 12m = + 2\sqrt {1 + {m^2}} \\\ \Rightarrow - 2\left( {1 + 6m} \right) = + 2\sqrt {1 + {m^2}} \\\
Squaring on both sides we get
⇒(−2(1+6m))2=(21+m2)2 ⇒4(1+36m2+12m)=4(1+m2) ⇒(1+36m2+12m)=(1+m2) ⇒1+36m2+12m−1−m2=0 ⇒35m2+12m=0 ⇒m(35m+12)=0 ⇒m=0,−1235  \Rightarrow {\left( { - 2\left( {1 + 6m} \right)} \right)^2} = {\left( {2\sqrt {1 + {m^2}} } \right)^2} \\\ \Rightarrow 4\left( {1 + 36{m^2} + 12m} \right) = 4\left( {1 + {m^2}} \right) \\\ \Rightarrow \left( {1 + 36{m^2} + 12m} \right) = \left( {1 + {m^2}} \right) \\\ \Rightarrow 1 + 36{m^2} + 12m - 1 - {m^2} = 0 \\\ \Rightarrow 35{m^2} + 12m = 0 \\\ \Rightarrow m\left( {35m + 12} \right) = 0 \\\ \Rightarrow m = 0, - \dfrac{{12}}{{35}} \\\
Therefore the correct option is a.

Note: A tangent to a circle is a straight line, in the plane of the circle, which touches the circle at only one point. The point is called the point of tangency or the point of contact.
The chord of the parabola which passes through the focus is called the focal chord.