Find the equation of the tangent to the curve | vedclass

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(A) The equation of the given curve is $y=\sqrt{3x-2}$.The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{

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$48x-24y=23$

Vedclass · Answer

(A) The equation of the given curve is $y=\sqrt{3x-2}$.The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{3}{2\sqrt{3x-2}}$.The equation of the given line is $4x-2y+5=0$,which can be written as $y=2x+\frac{5}{2}$.The slope of this line is $m=2$.Since the tangent is parallel to the line,their slopes must be equal:$\frac{3}{2\sqrt{3x-2}} = 2$$\Rightarrow \sqrt{3x-2} = \frac{3}{4}$Squaring both sides,we get $3x-2 = \frac{9}{16}$.$3x = \frac{9}{16} + 2 = \frac{41}{16}$$x = \frac{41}{48}$.Now,find the $y$-coordinate: $y = \sqrt{3(\frac{41}{48}) - 2} = \sqrt{\frac{41}{16} - 2} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.The point of tangency is $(\frac{41}{48}, \frac{3}{4})$.The equation of the tangent is $y - \frac{3}{4} = 2(x - \frac{41}{48})$.$y - \frac{3}{4} = 2x - \frac{41}{24}$.Multiply by $24$: $24y - 18 = 48x - 41$.$48x - 24y = 41 - 18 = 23$.Thus,the equation is $48x-24y=23$.

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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