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(C) For a parabola $y^2 = 4ax$,the equation of a tangent with slope $m$ is $y = mx + \frac{a}{m}$.For $y^2 = 4x$,we have $a = 1$,so the tangent is $y

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$x - y - 1 = 0$

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(C) For a parabola $y^2 = 4ax$,the equation of a tangent with slope $m$ is $y = mx + \frac{a}{m}$.For $y^2 = 4x$,we have $a = 1$,so the tangent is $y = mx + \frac{1}{m}$.If the point $P(h, k)$ lies on the tangent,then $k = mh + \frac{1}{m}$,which simplifies to $m^2h - mk + 1 = 0$.Let $m_1 = 	an 	heta_1$ and $m_2 = 	an 	heta_2$ be the slopes of the two tangents passing through $P(h, k)$.Then $m_1$ and $m_2$ are roots of the quadratic equation $m^2h - km + 1 = 0$.From the properties of roots,$m_1 + m_2 = \frac{k}{h}$ and $m_1 m_2 = \frac{1}{h}$.Given $	heta_1 + 	heta_2 = \frac{\pi}{4}$,we have $	an(	heta_1 + 	heta_2) = 	an(\frac{\pi}{4}) = 1$.Using the formula $	an(	heta_1 + 	heta_2) = \frac{m_1 + m_2}{1 - m_1 m_2} = 1$.Substituting the values: $\frac{k/h}{1 - 1/h} = 1$.$\frac{k/h}{(h-1)/h} = 1 \Rightarrow \frac{k}{h-1} = 1$.Thus,$k = h - 1$,or $h - k - 1 = 0$.The locus of $P(h, k)$ is $x - y - 1 = 0$.

$A$ pair of tangents is drawn from an external point $P$ to the parabola $y^2 = 4x$. If $\theta_1$ and $\theta_2$ are the angles made by the tangents with the $x$-axis such that $\theta_1 + \theta_2 = \frac{\pi}{4}$,find the locus of $P$.

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