Question

Find the equation of the tangent to the curve which is parallel to the line 4x − 2y + 5 = 0

Answer 1

The equation of the given curve is $y=\sqrt{3x-2}$.

The slope of the tangent to the given 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

4x − 2y + 5 = 0

∴ y=2x+$\frac{5}{2}$ (which is of the form y=mx+c)

∴The slope of the line = 2 Now, the tangent to the given curve is parallel to the line 4x − 2y − 5 = 0 if the slope of the tangent is equal to the slope of the line.

$\frac{3}{2}\sqrt{3x-2}$

3x-2=$\frac{3}{4}$

3x-2=$\frac{9}{16}$

3x=$\frac{9}{16}$+2=$\frac{41}{16}$+2=$\frac{41}{16}$

x=$\frac{41}{48}$

when x=$\frac{41}{48}$, y=$\sqrt{3(\frac{41}{48})}$-2=$\sqrt{\frac{41}{16}}-2\sqrt41-\frac{32}{16}$=$\sqrt{\frac{9}{16}}$=3/4.

∴The equation of the tangent passing through the point is given by,

=48x-24y=23

Hence, the equation of the required tangent is 48-24y=23.