Question

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

Answer 1

Hint: Find the differentiation of the curve given in the question from that we can get the slope of the tangent. By using the slope of a given line, we can get the slope of the tangent. By these both we can get the point of contact, from the point of contact and slope we can achieve the equation of line.

Complete step-by-step answer:

Slope of tangent to curve y at (a, b) = \[{{\left. \dfrac{dy}{dx} \right|}_{\left( x=a,y=b \right)}}\]

Let (h, k) be the point on the curve from tangent to be taken.

By basic knowledge of differentiation, we can say that:

Slope of tangent at (a, b) = \[{{\left. \dfrac{dy}{dx} \right|}_{\left( x=a,y=b \right)}}\]

Given in equation, equation of curve y to be:

$y=\sqrt{3x-2}$

By differentiating with respect to $x$ on both sides, we get:

$\dfrac{dy}{dx}=\dfrac{d}{dx}\sqrt{3x-2}$

By basic knowledge of differentiation, we can say:

$\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}};\text{ }\dfrac{d}{dx}\left( kx \right)=k\text{ (k = constant)}$

By applying above equations here, we convert it into

$\dfrac{dy}{dx}=\dfrac{3}{2\sqrt{3x-2}}$

As a point we assumed as (h, k). Substitute it to get the slope of tangent at (h, k).

Slope = \[\dfrac{3}{2\sqrt{3h-2}}...........(i)\]

Given tangent is parallel to the line $4x-2y+5$

By above condition, we can derive an equation that:

Slope of tangent = Slope of $4x-2y+5$

Slope of $4x-2y+5=0$

By adding $-4x-5$ on both sides of line, we get:

$-2y=-4x-5$

By dividing with -2 on both sides of line, we get:

$y=2x+\dfrac{5}{2}$

By comparing it with the slope from $y=mx+c$ , we get m = 2

Slope of line \[4x-2y+5=0\] is 2

By above equation we can say that slope:

Slope of tangent = Slope of line

Slope of tangent = 2

By substituting this into the equation (i), we get:

$2=\dfrac{3}{2\sqrt{3h-2}}$

By cross multiplying in this equation, we get:

$\sqrt{3h-2}=\dfrac{3}{4}$

By squaring on both sides of equation, we get

\[3h-2=\dfrac{9}{16}\]

By adding 2 on both sides of equation, we get:

$3h=\dfrac{41}{16}$

By dividing with 3 on both sides we get:

$h=\dfrac{41}{48}$

As we know (h, k) lies on the curve. So, by substituting point into curve

$k=\sqrt{3h-2}=\sqrt{3\left( \dfrac{41}{48} \right)-2}=\sqrt{\dfrac{9}{16}}$

$k=\dfrac{3}{4}$

So, point will be $\left( \dfrac{41}{48},\dfrac{3}{4} \right)$ slope is 2

Line equation with slope m through (a, b)

$y-b=m\left( x-a \right)$

By using above, we get line equation:

$y-\dfrac{3}{4}=2\left( x-\dfrac{41}{48} \right)$

By taking Least common multiple, we get:

$\dfrac{4y-3}{4}=2\left( \dfrac{48x-41}{48} \right)$

By cross multiplication of terms, we get:

$24y-18=48x-41$

By simplifying the equation, we get:

$48x-24y=23$

Therefore, the equation of tangent $48x-24y-23=0$


Note: Be careful while calculating slope as it is parallel to line given, both are equal.