13.0 Circle through co-normal points

  Parabola

Let us assume that the three normals drawn from a given point $P(\alpha ,\beta )$ to the parabola ${y^2} = 4ax$ and the coordinates of three points on the parabola be $A(a{m_1}^2, - 2a{m_1})$, $B(a{m_2}^2, - 2a{m_2})$, $C(a{m_3}^2, - 2a{m_3})$.

The equation of normal to the parabola in slope form is $$y = mx - 2am - a{m^3}$$

If it passes through point $P(\alpha ,\beta )$, then point satisfies the equation. We get

$$\beta = m\alpha - 2am - a{m^3}$$ $$a{m^3} + m\left( {2a - \alpha } \right) + \beta = 0...(1)$$

and roots of the equation are ${m_1}$, ${m_2}$ and ${m_3}$ such that $${m_1} + {m_2} + {m_3} = 0...(2)$$ $${m_1}{m_2} + {m_2}{m_3} + {m_3}{m_1} = \frac{{2a - \alpha }}{a}...(3)$$ and $${m_1}{m_2}{m_3} = \frac{{ - \beta }}{a}...(4)$$

Let the equation of the circle passing through $A$, $B$ and $C$ be $${x^2} + {y^2} + 2gx + 2fy + c = 0...(5)$$

If a point $(a{m^2}, - 2am)$ lies on it then,

$${\left( {a{m^2}} \right)^2} + {\left( { - 2am} \right)^2} + 2g\left( {a{m^2}} \right) + 2f\left( { - 2am} \right) + c = 0$$

or, $${a^2}{m^4} + \left( {4{a^2} + 2ag} \right){m^2} - 4afm + c = 0...(6)$$

This is a biquadratic equation in $m$. Hence there are four values of $m$, say ${m_1}$, ${m_2}$, ${m_3}$ and ${m_4}$ such that the circle pass through the points $A(a{m_1}^2, - 2a{m_1})$, $B(a{m_2}^2, - 2a{m_2})$, $C(a{m_3}^2, - 2a{m_3})$ and $D(a{m_4}^2, - 2a{m_4})$.

Therefore, $${m_1} + {m_2} + {m_3} + {m_4} = 0...(7)$$ or, $$0 + {m_4} = 0$$ $${m_4} = 0$$

Therefore, the coordinates of point $D(a{m_4}^2, - 2a{m_4})$$ = (0,0)$

Thus the circle passes through the vertex of the parabola ${y^2} = 4ax$. Satisfy $(0,0)$ in equation $(5)$, we get $$0+0+0+0+c=0$$ $$c=0$$

Now, equation $(6)$ becomes $$am\left( {a{m^3} + \left( {4a + 2g} \right)m - 4f} \right) = 0$$ $$a{m^3} + \left( {4a + 2g} \right)m - 4f = 0...(8)$$

Equation $(1)$ and $(8)$ are identical. On comparing we get, $$1 = \frac{{4a + 2g}}{{2a - \alpha }} = \frac{{ - 4f}}{\beta }$$

or, $$2g = - \left( {2a + \alpha } \right)$$ $$2f = - \frac{\beta }{2}$$

Therefore, the equation of required circle is ${x^2} + {y^2} - \left( {2a + \alpha } \right){\text{}}x - \frac{\beta }{2}y = 0$.