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.