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

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(A) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.Let the point of intersection be $(h, k)$. Since th

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$y = ac$

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(A) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.Let the point of intersection be $(h, k)$. Since the tangent passes through $(h, k)$,we have $k = mh + \frac{a}{m}$.Rearranging this gives the quadratic equation in $m$: $m^2h - mk + a = 0$.Let $m_1 = 	an 	heta_1$ and $m_2 = 	an 	heta_2$ be the roots of this equation.From the properties of roots,$m_1 + m_2 = \frac{k}{h}$ and $m_1 m_2 = \frac{a}{h}$.Given $\cot 	heta_1 + \cot 	heta_2 = c$,we can write this as $\frac{1}{	an 	heta_1} + \frac{1}{	an 	heta_2} = c$.This simplifies to $\frac{	an 	heta_1 + 	an 	heta_2}{	an 	heta_1 	an 	heta_2} = c$.Substituting the values of the sum and product of roots: $\frac{k/h}{a/h} = c$.This simplifies to $\frac{k}{a} = c$,which implies $k = ac$.Replacing $(h, k)$ with $(x, y)$,the locus is $y = ac$.

The tangents to the parabola $y^2 = 4ax$ make angles $\theta_1$ and $\theta_2$ with the positive $x$-axis. If $\cot \theta_1 + \cot \theta_2 = c$,then the locus of their point of intersection is

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