202211231536 Solution to 1986-CE-AMATH-I-4

Find the equation of the tangent to the curve

$x^2+xy+y^2=7$

at the point $(2,1)$ .

Roughwork.

$\begin{aligned} x^2+xy+y^2-7 & = 0 \\ 2x\,\mathrm{d}x+x\,\mathrm{d}y+y\,\mathrm{d}x + 2y\,\mathrm{d}y & = 0 \\ 2x + x\frac{\mathrm{d}y}{\mathrm{d}x}+y+2y\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ -\frac{2x+y}{x+2y} & = y'(x) \\ \end{aligned}$

The slope of tangent at $(x=2,y=1)$ is

$y'(x)|_{(2,1)}=\displaystyle{-\frac{2(2)+(1)}{(2)+2(1)}=-\frac{5}{4}}$ .

and its equation

$\begin{aligned} & \quad\enspace \frac{y-1}{x-2} = -\frac{5}{4} \\ & \Rightarrow 5x + 4y = 14 \\ \end{aligned}$

M	T	W	T	F	S	S
	1	2	3	4	5	6
7	8	9	10	11	12	13
14	15	16	17	18	19	20
21	22	23	24	25	26	27
28	29	30