Through the vertex O of the parabola y^2 = 4a | vedclass

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(A) Let the coordinates of $P$ be $(at_1^2, 2at_1)$ and $Q$ be $(at_2^2, 2at_2)$.The slope of the tangent at $P$ is $m_1 = \frac{1}{t_1} = 	an 	heta

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$-2 	an \phi$

Vedclass · Answer

(A) Let the coordinates of $P$ be $(at_1^2, 2at_1)$ and $Q$ be $(at_2^2, 2at_2)$.The slope of the tangent at $P$ is $m_1 = \frac{1}{t_1} = 	an 	heta_1$,so $\cot 	heta_1 = t_1$.The slope of the tangent at $Q$ is $m_2 = \frac{1}{t_2} = 	an 	heta_2$,so $\cot 	heta_2 = t_2$.The circle with diameter $OP$ has equation $x(x - at_1^2) + y(y - 2at_1) = 0$,i.e.,$x^2 + y^2 - at_1^2x - 2at_1y = 0$.The circle with diameter $OQ$ has equation $x^2 + y^2 - at_2^2x - 2at_2y = 0$.Subtracting these equations gives the common chord $OR$: $a(t_1^2 - t_2^2)x + 2a(t_1 - t_2)y = 0$.Since $t_1 
eq t_2$,we divide by $a(t_1 - t_2)$ to get $(t_1 + t_2)x + 2y = 0$.The slope of $OR$ is $	an \phi = -\frac{t_1 + t_2}{2}$.Thus,$\cot 	heta_1 + \cot 	heta_2 = t_1 + t_2 = -2 	an \phi$.

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