Question

Find the value of \(c\) such that the line \(y = \frac{3}{2}x + 6\) is tangent to the curve \(y = c\sqrt x \).

Short answer

The value of \(c\) is \(6\).

Step-by-step solution

Tangent to a curve

A tangent to any curve is a straight line which passes through the curve by touching it any point and is expressed in the form similar to the equation of the line passing through any point \(\left( {x,y} \right)\).

Evaluating equation of parabola.

Thegiven equation is \(y = c\sqrt x \).

Differentiating with respect to\(x\), we have:

\(y' = \;\frac{c}{{2\sqrt x }}\)

Now, equation of tangent line is \(y = \frac{3}{2}x + 6\)has slope \(\frac{3}{2}\), so at \(x = a\), we have:

\(\begin{aligned}y'\left| {_{x = a}} \right. &= \frac{3}{2}\\\frac{c}{{2\sqrt a }} &= \frac{3}{2}\\c &= 3\sqrt a \end{aligned}\)

Evaluating the tangent equation.

Now, tangent equation at \(x = a\)will be:

\(\begin{aligned}c\sqrt a &= \frac{3}{2}a + 6\\3a &= \frac{3}{2}a + 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{as}}\,\,c = 3\sqrt a } \right)\\3a - \frac{3}{2}a &= 6\\\frac{{6a - 3a}}{2} &= 6\\3a &= 12\\a &= 4\end{aligned}\)

So, we have:

\(\begin{aligned}c &= 3\sqrt a \\ &= 3\sqrt 4 \\ &= 6\end{aligned}\)

Hence, this is the required answer.