Find the equation of the tangent to the curve \(y=\sqrt{3x-2}\) which

1.	Find the equation of the tangent to the curve \(y=\sqrt{3x-2}\) which is parallel to the line 4x – 2y + 5 = 0.
Answer» finding the slope of the tangent by differentiating the curve \(\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}\) equation of tangent is given by y – y₁ = m(tangent)(x – x₁) now comparing the slope of a tangent with the given equation m(tangent) = 2 \(\frac{3}{2\sqrt{3x-2}}=2\) \(\frac{9}{16}=3x-2\) \(x=\frac{41}{48}\) since this point lies on the curve, we can find y by substituting x \(y=\sqrt{\frac{41}{16}-2}\) \(y=\frac{3}{4}\) therefore, the equation of the tangent is \(y-\frac{3}{4}=2(x-\frac{41}{48})\)

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

Answer»

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

m(tangent) = 2

\(\frac{3}{2\sqrt{3x-2}}=2\)

\(\frac{9}{16}=3x-2\)

\(x=\frac{41}{48}\)

since this point lies on the curve, we can find y by substituting x

\(y=\sqrt{\frac{41}{16}-2}\)

\(y=\frac{3}{4}\)

therefore, the equation of the tangent is

\(y-\frac{3}{4}=2(x-\frac{41}{48})\)