If the line lx + my + n = 0 is a tangent to t | vedclass

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(C) The equation of a tangent to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is given by $ty = x + at^2$,which can be rewritten as $x - ty + a

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$A$ parabola

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(C) The equation of a tangent to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ is given by $ty = x + at^2$,which can be rewritten as $x - ty + at^2 = 0$. Given that the line $lx + my + n = 0$ is a tangent to the parabola,it must be identical to the equation $x - ty + at^2 = 0$. Comparing the coefficients of the two equations,we have: $\frac{1}{l} = \frac{-t}{m} = \frac{at^2}{n}$ From $\frac{1}{l} = \frac{-t}{m}$,we get $t = -\frac{m}{l}$. From $\frac{1}{l} = \frac{at^2}{n}$,we get $t^2 = \frac{n}{al}$. Substituting $t = -\frac{m}{l}$ into $t^2 = \frac{n}{al}$,we get: $(-\frac{m}{l})^2 = \frac{n}{al} \implies \frac{m^2}{l^2} = \frac{n}{al} \implies m^2 = \frac{nl}{a}$. The point of contact is $(at^2, 2at)$. Let $(x_1, y_1)$ be the point of contact. Then $x_1 = at^2 = a(\frac{n}{al}) = \frac{n}{l}$ and $y_1 = 2at = 2a(-\frac{m}{l}) = -\frac{2am}{l}$. Since the point of contact $(x_1, y_1)$ lies on the parabola $y^2 = 4ax$,the locus of the point of contact is the parabola itself.

If the line $lx + my + n = 0$ is a tangent to the parabola $y^2 = 4ax$,then the locus of its point of contact is

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