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Question: A circle is described on a focal chord as diameter; If m be the tangent of the inclination of the ch...

A circle is described on a focal chord as diameter; If m be the tangent of the inclination of the chord to the axis, prove that the equation on the circle is
x2+y2−2ax(1+2m2)−4aym−3a2=0{x^2} + {y^2} - 2ax\left( {1 + \dfrac{2}{{{m^2}}}} \right) - \dfrac{{4ay}}{m} - 3{a^2} = 0

Explanation

Solution

Hint: Simply by applying the parametric equation of parabola and equation of circle we can easily solve the question.

Complete step-by-step answer:
Let the end points of the focal chord be P(at12,2at1)P\left( {at_1^2,2a{t_1}} \right) and Q(at22,2at2)Q\left( {at_2^2,2a{t_2}} \right)
Slope of PQ = 2a(t2−t1)a(t2−t1)(t2+t1) \dfrac{{2a\left( {{t_2} - {t_1}} \right)}}{{a\left( {{t_2} - {t_1}} \right)\left( {{t_2} + {t_1}} \right)}}
This gives us,
2at2−2at1at22−at12\dfrac{{2a{t_2} - 2a{t_1}}}{{at_2^2 - at_1^2}} = m
⇒2a(t2−t1)a(t2−t1)(t2+t1)\Rightarrow \dfrac{{2a\left( {{t_2} - {t_1}} \right)}}{{a\left( {{t_2} - {t_1}} \right)\left( {{t_2} + {t_1}} \right)}} = m
⇒\Rightarrow t1+t2=2m{t_1} + {t_2} = \dfrac{2}{m} Equation (1)
As you know that parametric form of equation of circle is
(x−at12)(x−at22)+(y−2at1)(y−2at2)=0\left( {x - at_1^2} \right)\left( {x - at_2^2} \right) + \left( {y - 2a{t_1}} \right)\left( {y - 2a{t_2}} \right) = 0
Simplifying further, we get,
⇒\Rightarrow x2−ax(t12+t22)+a2t12t22+y2−2ay(t1+t2)+4a2t1t2=0{x^2} - ax\left( {t_1^2 + t_2^2} \right) + {a^2}t_1^2t_2^2 + {y^2} - 2ay\left( {{t_1} + {t_2}} \right) + 4{a^2}{t_1}{t_2} = 0
⇒\Rightarrow x2−ax((t12+t22)2−2t1t2)+a2t12t22+y2−2ay(t1+t2)+4a2t1t2=0{x^2} - ax\left( {{{\left( {t_1^2 + t_2^2} \right)}^2} - 2{t_1}{t_2}} \right) + {a^2}t_1^2t_2^2 + {y^2} - 2ay\left( {{t_1} + {t_2}} \right) + 4{a^2}{t_1}{t_2} = 0
As we know that for focal chord, t1t2=−1{t_1}{t_2} = - 1
⇒\Rightarrow x2−ax((t12+t22)2+2)+a2+y2−2ay(t1+t2)−4a2=0{x^2} - ax\left( {{{\left( {t_1^2 + t_2^2} \right)}^2} + 2} \right) + {a^2} + {y^2} - 2ay\left( {{t_1} + {t_2}} \right) - 4{a^2} = 0
⇒\Rightarrow x2−ax((t12+t22)2+2)+y2−2ay(t1+t2)−3a2=0{x^2} - ax\left( {{{\left( {t_1^2 + t_2^2} \right)}^2} + 2} \right) + {y^2} - 2ay\left( {{t_1} + {t_2}} \right) - 3{a^2} = 0
Now if we substitute Equation (1) in above equation
We get,
⇒\Rightarrow x2−ax((2m)2+2)+y2−2ay(2m)−3a2=0{x^2} - ax\left( {{{\left( {\dfrac{2}{m}} \right)}^2} + 2} \right) + {y^2} - 2ay\left( {\dfrac{2}{m}} \right) - 3{a^2} = 0
⇒x2−ax(4m2+2)+y2−2ay(2m)−3a2=0\Rightarrow {x^2} - ax\left( {\dfrac{4}{{{m^2}}} + 2} \right) + {y^2} - 2ay\left( {\dfrac{2}{m}} \right) - 3{a^2} = 0
⇒x2−2ax(2m2+1)+y2−(4aym)−3a2=0\Rightarrow {x^2} - 2ax\left( {\dfrac{2}{{{m^2}}} + 1} \right) + {y^2} - \left( {\dfrac{{4ay}}{m}} \right) - 3{a^2} = 0
Hence Proved.

Note: To solve this question firstly you must be clear with the concept of parametric equation of parabola and equation of circle that can be written using two points on the circle. If you use these both then you can easily solve the question.