Find the equation of the tangent to the curve | vedclass

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(B) The slope of the line $4x - 2y + 5 = 0$ is $m = 2$. Since the tangent is parallel to this line,its slope must also be $2$.Given the curve $y = \sq

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$48x - 24y - 23 = 0$

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(B) The slope of the line $4x - 2y + 5 = 0$ is $m = 2$. Since the tangent is parallel to this line,its slope must also be $2$.Given the curve $y = \sqrt{3x - 2}$,we find the derivative:$\frac{dy}{dx} = \frac{1}{2\sqrt{3x - 2}} 	imes 3 = \frac{3}{2\sqrt{3x - 2}}$.Setting the slope equal to $2$:$\frac{3}{2\sqrt{3x - 2}} = 2$$3 = 4\sqrt{3x - 2}$$9 = 16(3x - 2)$$9 = 48x - 32$$48x = 41 \Rightarrow x = \frac{41}{48}$.Now,find the $y$-coordinate:$y = \sqrt{3(\frac{41}{48}) - 2} = \sqrt{\frac{41}{16} - \frac{32}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.The point of tangency is $(\frac{41}{48}, \frac{3}{4})$.The equation of the tangent is $y - y_1 = m(x - x_1)$:$y - \frac{3}{4} = 2(x - \frac{41}{48})$$y - \frac{3}{4} = 2x - \frac{41}{24}$$y = 2x - \frac{41}{24} + \frac{18}{24}$$y = 2x - \frac{23}{24}$$24y = 48x - 23$$48x - 24y - 23 = 0$.

Find the equation of the tangent to the curve $y = \sqrt{3x - 2}$ which is parallel to the line $4x - 2y + 5 = 0$.

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