202212231530 Solution to 1990-CE-AMATH-I-7

Given the curve $C: x^2+4xy+5y^2=1$ , find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ . Hence find the equations of the two tangents to $C$ which are parallel to the line $\displaystyle{y=-\frac{1}{2}x}$ .

Roughwork.

For the curve,

$\begin{aligned} &\quad\enspace x^2+4xy+5y^2 = 1 \\ &\Rightarrow 2x\,\mathrm{d}x + 4(x\,\mathrm{d}y + y\,\mathrm{d}x) + 10y\,\mathrm{d}y = 0 \\ &\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}= -\frac{x+2y}{2x+5y} \\ \end{aligned}$

and for the line,

$\begin{aligned} &\quad\enspace y = -\frac{1}{2}x \\ &\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2} \\ \end{aligned}$

equating them,

$\begin{aligned} &\quad\enspace -\frac{x+2y}{2x+5y} = -\frac{1}{2} \\ &\Rightarrow \begin{pmatrix} x \\ y \end{pmatrix} = \bigg\{\begin{pmatrix} 1 \\ 0 \end{pmatrix} , \begin{pmatrix} -1 \\ 0 \end{pmatrix}\bigg\} \\ \end{aligned}$

are points of contact at which the tangents

$\begin{aligned} &\quad\enspace \frac{y-(\pm 1)}{x-(0)} = -\frac{1}{2} \\ &\Rightarrow \big\{ x + 2y \pm 2 = 0 \big\} \\ \end{aligned}$

are drawn.

This problem is not to be attempted.

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	31