The tangents to the parabola y^2 = 4ax from a | vedclass

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(D) Let the point of contact of the tangents from $P$ to the parabola $y^2 = 4ax$ be $A(at_1^2, 2at_1)$ and $B(at_2^2, 2at_2)$.The point of intersecti

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$y = bx$

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(D) Let the point of contact of the tangents from $P$ to the parabola $y^2 = 4ax$ be $A(at_1^2, 2at_1)$ and $B(at_2^2, 2at_2)$.The point of intersection $P$ of the tangents at $A$ and $B$ is given by $P(at_1t_2, a(t_1 + t_2))$.Let $P = (x, y)$,so $x = at_1t_2$ and $y = a(t_1 + t_2)$.The slope of the tangent at $A(at_1^2, 2at_1)$ is $m_1 = 	an 	heta_1 = \frac{1}{t_1}$.The slope of the tangent at $B(at_2^2, 2at_2)$ is $m_2 = 	an 	heta_2 = \frac{1}{t_2}$.Given that $	an 	heta_1 + 	an 	heta_2 = b$,we have $\frac{1}{t_1} + \frac{1}{t_2} = b$.This simplifies to $\frac{t_1 + t_2}{t_1t_2} = b$.Substituting the coordinates of $P$,we get $\frac{y/a}{x/a} = b$,which implies $\frac{y}{x} = b$.Therefore,$y = bx$.

The tangents to the parabola $y^2 = 4ax$ from an external point $P$ make angles $\theta_1$ and $\theta_2$ with the axis of the parabola,such that $\tan \theta_1 + \tan \theta_2 = b$,where $b$ is a constant. Then $P$ lies on

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