A tangent to the parabola y^2 = 8x makes an a | vedclass

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(C) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The slope of the line $y = 3x + 5$ is $m_1 = 3$.Let the slope of the ta

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$2x + y + 1 = 0$

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(C) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The slope of the line $y = 3x + 5$ is $m_1 = 3$.Let the slope of the tangent be $m$. Since the angle between the tangent and the line is $45^\circ$,we have $	an 45^\circ = |\frac{m - m_1}{1 + m \cdot m_1}|$.$1 = |\frac{m - 3}{1 + 3m}|$.This gives two cases:Case $1$: $\frac{m - 3}{1 + 3m} = 1 \implies m - 3 = 1 + 3m \implies -2m = 4 \implies m = -2$.Case $2$: $\frac{m - 3}{1 + 3m} = -1 \implies m - 3 = -1 - 3m \implies 4m = 2 \implies m = \frac{1}{2}$.The equation of a tangent to $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.For $m = -2$: $y = -2x + \frac{2}{-2} \implies y = -2x - 1 \implies 2x + y + 1 = 0$.For $m = \frac{1}{2}$: $y = \frac{1}{2}x + \frac{2}{1/2} \implies y = \frac{1}{2}x + 4 \implies x - 2y + 8 = 0$.Comparing with the given options,$2x + y + 1 = 0$ is the correct equation.

$A$ tangent to the parabola $y^2 = 8x$ makes an angle of $45^\circ$ with the straight line $y = 3x + 5$. Find the equation of the tangent.

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