Question

Given that the gradient of the curve $y = mx^{2} - \frac{1}{nx}$ at point N (-1,9) is -9. Find the value of m and n

Answer 1

Step-by-step explanation:

We have,

$y = m {x}^{2} - \frac{1}{nx}$

$\implies \: \frac{dy}{dx} = 2mx + \frac{1}{n {x}^{2} }$

$\implies \: \frac{dy}{dx}_{(x = - 1 \: y = 9)} = - 2m + \frac{1}{n}$

Also,

$9 = m + \frac{1}{n}$

$\implies \frac{1}{n} = 9 - m$

$\implies - 9 = - 2m + 9 - m$

$\implies3m = 18\implies \: m = 6$

$\therefore \: n = \frac{1}{3}$

Answer 2

The curve is given by,

$\longrightarrow y=mx^2-\dfrac{1}{nx}$

At $(x,\ y)=(-1,\ 9),$

$\longrightarrow 9=m(-1)^2-\dfrac{1}{n(-1)}$

$\longrightarrow 9=m+\dfrac{1}{n}$

$\longrightarrow m=9-\dfrac{1}{n}\quad\quad\dots(1)$

The slope of tangent to the curve is given by,

$\longrightarrow s=\dfrac{dy}{dx}$

$\longrightarrow s=\dfrac{d}{dx}\left(mx^2-\dfrac{1}{nx}\right)$

$\longrightarrow s=2mx+\dfrac{1}{nx^2}$

At $(x,\ y)=(-1,\ 9),$

$\longrightarrow s=-9$

$\longrightarrow2m(-1)+\dfrac{1}{n(-1)^2}=-9$

$\longrightarrow-2m+\dfrac{1}{n}=-9$

Putting value of $m$ from (1),

$\longrightarrow-2\left(9-\dfrac{1}{n}\right)+\dfrac{1}{n}=-9$

$\longrightarrow-18+\dfrac{3}{n}=-9$

$\longrightarrow n=\dfrac{1}{3}$

Putting value of $n$ in (1),

$\longrightarrow m=9-\dfrac{1}{\left(\dfrac{1}{3}\right)}$

$\longrightarrow m=6$

Hence,

$\longrightarrow\underline{\underline{\left(m,\ n\right)=\left(6,\ \dfrac{1}{3}\right)}}$