If a focal chord of the parabola y^2 = 4x mak | vedclass

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(B) For the parabola $y^2 = 4ax$,where $a = 1$,the focal chord makes an angle $	heta = 45^{\circ}$ with the $X$-axis. The slope of the focal chord is

Vedclass · Accepted Answer

$m^2 + 2m - 1 = 0$

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(B) For the parabola $y^2 = 4ax$,where $a = 1$,the focal chord makes an angle $	heta = 45^{\circ}$ with the $X$-axis. The slope of the focal chord is $m_c = 	an(45^{\circ}) = 1$.Let the ends of the focal chord be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$. Since it is a focal chord,$t_1 t_2 = -1$.The slope of the chord is $m_c = \frac{2}{t_1 + t_2} = 1$,which implies $t_1 + t_2 = 2$.The slope of the normal at any point $t$ on the parabola $y^2 = 4ax$ is given by $m = -t$.Thus,the slopes of the normals at the ends are $m_1 = -t_1$ and $m_2 = -t_2$.We need to find the equation satisfied by $m_1$ and $m_2$. The sum of the roots is $m_1 + m_2 = -(t_1 + t_2) = -2$.The product of the roots is $m_1 m_2 = (-t_1)(-t_2) = t_1 t_2 = -1$.The quadratic equation with roots $m_1$ and $m_2$ is $m^2 - (m_1 + m_2)m + m_1 m_2 = 0$.Substituting the values,we get $m^2 - (-2)m + (-1) = 0$,which simplifies to $m^2 + 2m - 1 = 0$.

If a focal chord of the parabola $y^2 = 4x$ makes an angle $45^{\circ}$ with the positive $X$-axis,then the slopes of the normals drawn at the ends of the focal chord will satisfy the equation:

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