Find the values of m and n in a function given the tangent line at a point

Question

The question is: If $y = m\sqrt{1-nx}$, (where m and n are constants) has a tangent line $6x + 2y = 10$ at $x = -1$. Find the values of m and n.

Attempted solution:

I isolated y in both equations and set them equal to get one equation for m in terms of n:

$5 - 3x = m\sqrt{1-nx}$

$m = \frac{8}{\sqrt{1+n}}$

To get the second equation I set the derivatives of the two original equations equal and solved for m:

$-3 = \frac{1}{2}m(1+n)^{\frac{-1}{2}}$

$m = -6\sqrt{1+n}$

But when I try to solve for m, I don't get the right answer. Also, If I put the correct solutions for n and m into my second equation it results in an untrue statement.

How do I find the correct second equation?

The solutions are: $m = 4, n = 3$

Answer 1

As you differentiate it, the gradient should be

$$-3 =- \frac{mn}2 (1+n)^{-\frac12}$$

Your mistake is you forgot to apply chain rule and omitted a $-n$ factor on the RHS.

From the first equation, we have

$$m = \frac{8}{\sqrt{1+n}}$$

Hence we have

$$-3=-\frac{4n}{1+n}$$

and we have

$$3+3n=4n$$

Hence $n=3$.

Answer 2

When you take the derivative of $5 - 3x = m\sqrt{1-nx}$ wrt $x$ you forgot sth in RHS:

That is it should be $-3 = \frac{1}{2}m(1-nx)^{\frac{-1}{2}}(-n)$
then you get $-3 = \frac{1}{2}m(1+n)^{\frac{-1}{2}}(-n)\\ \iff 6=mn(1+n)^{\frac{-1}{2}} \iff 6=\frac {m^2 n} {8} \iff 48=\frac {64n}{1+n} \iff n=3$