Question

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
${x^2} + {y^2} - 2ax\left( {1 + \dfrac{2}{{{m^2}}}} \right) - \dfrac{{4ay}}{m} - 3{a^2} = 0$

Answer 1

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 $P\left( {at_1^2,2a{t_1}} \right)$ and \[Q\left( {at_2^2,2a{t_2}} \right)\]
Slope of PQ = $ \dfrac{{2a\left( {{t_2} - {t_1}} \right)}}{{a\left( {{t_2} - {t_1}} \right)\left( {{t_2} + {t_1}} \right)}}$
This gives us,
$\dfrac{{2a{t_2} - 2a{t_1}}}{{at_2^2 - at_1^2}}$ = m
$ \Rightarrow \dfrac{{2a\left( {{t_2} - {t_1}} \right)}}{{a\left( {{t_2} - {t_1}} \right)\left( {{t_2} + {t_1}} \right)}}$ = m
$ \Rightarrow $ ${t_1} + {t_2} = \dfrac{2}{m}$ Equation (1)
As you know that parametric form of equation of circle is
\[\left( {x - at_1^2} \right)\left( {x - at_2^2} \right) + \left( {y - 2a{t_1}} \right)\left( {y - 2a{t_2}} \right) = 0\]
Simplifying further, we get,
$\Rightarrow$ \[{x^2} - ax\left( {t_1^2 + t_2^2} \right) + {a^2}t_1^2t_2^2 + {y^2} - 2ay\left( {{t_1} + {t_2}} \right) + 4{a^2}{t_1}{t_2} = 0\]
$\Rightarrow$ \[{x^2} - ax\left( {{{\left( {t_1^2 + t_2^2} \right)}^2} - 2{t_1}{t_2}} \right) + {a^2}t_1^2t_2^2 + {y^2} - 2ay\left( {{t_1} + {t_2}} \right) + 4{a^2}{t_1}{t_2} = 0\]
As we know that for focal chord, ${t_1}{t_2} = - 1$
$\Rightarrow$ \[{x^2} - ax\left( {{{\left( {t_1^2 + t_2^2} \right)}^2} + 2} \right) + {a^2} + {y^2} - 2ay\left( {{t_1} + {t_2}} \right) - 4{a^2} = 0\]
$\Rightarrow$ \[{x^2} - ax\left( {{{\left( {t_1^2 + t_2^2} \right)}^2} + 2} \right) + {y^2} - 2ay\left( {{t_1} + {t_2}} \right) - 3{a^2} = 0\]
Now if we substitute Equation (1) in above equation
We get,
$\Rightarrow$ ${x^2} - ax\left( {{{\left( {\dfrac{2}{m}} \right)}^2} + 2} \right) + {y^2} - 2ay\left( {\dfrac{2}{m}} \right) - 3{a^2} = 0$
$ \Rightarrow {x^2} - ax\left( {\dfrac{4}{{{m^2}}} + 2} \right) + {y^2} - 2ay\left( {\dfrac{2}{m}} \right) - 3{a^2} = 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.