From an external point P(h, k),a pair of tang | vedclass

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(C) The equation of a tangent to the parabola $y^2 = 4x$ with slope $m$ is $y = mx + \frac{1}{m}$.Since the tangent passes through $P(h, k)$,we have $

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$x - y - 1 = 0$

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(C) The equation of a tangent to the parabola $y^2 = 4x$ with slope $m$ is $y = mx + \frac{1}{m}$.Since the tangent passes through $P(h, k)$,we have $k = mh + \frac{1}{m}$,which simplifies to $m^2h - mk + 1 = 0$.Let $m_1 = 	an 	heta_1$ and $m_2 = 	an 	heta_2$ be the roots of this quadratic equation.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-1} = 1 \Rightarrow k = h - 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $y = x - 1$,or $x - y - 1 = 0$.

From an external point $P(h, k)$,a pair of tangent lines are drawn to the parabola $y^2 = 4x$. If $\theta_1$ and $\theta_2$ are the inclinations of these tangents with the $x$-axis such that $\theta_1 + \theta_2 = \frac{\pi}{4}$,then the locus of $P$ is:

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