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 | 1 | initial version | posted 2014-07-16 20:04:41 -0600  |
Given the system of equations:
$$ 2 - 2 \lambda x = 0 $$ $$ 4 - 2 \lambda x = 0 $$ $$ x^2 + y^2 = 20 $$
We can simply solve for $x$ and $y$ in terms of $\lambda$ from the first two equations and plug it in to the third equation.
Solving for $x$: $$ 2 - 2 \lambda x = 0 $$ $$ 2 \lambda x = 2 $$ $$ x = \frac{1}{\lambda} $$
Solving for $y$: $$ 4 - 2 \lambda y = 0 $$ $$ 2 \lambda y = 4 $$ $$ y = \frac{2}{\lambda} $$
Plugging in to the third equation: $$ \left(\frac{1}{\lambda}\right)^2 + \left(\frac{2}{\lambda}\right)^2 = 20 $$ $$ \frac{1}{\lambda^2} + \frac{4}{\lambda^2} = 20 $$ $$ \frac{5}{\lambda^2} = 20 $$ $$ \lambda^2 = \frac{5}{20} = \frac{1}{4} $$ $$ \lambda = \pm \frac{1}{2} $$
This means that there are $2$ cases to examine. When $\lambda = -\frac{1}{2}$: $$ x = \frac{1}{-\frac{1}{2}} = -2 $$ $$ y = \frac{2}{-\frac{1}{2}} = -4 $$
When $\lambda = \frac{1}{2}$: $$ x = \frac{1}{\frac{1}{2}} = 2 $$ $$ y = \frac{2}{\frac{1}{2}} = 4 $$
The two points that solve this system of equations are $(-2, -4)$ and $(2, 4)$.
| | 2 | No.2 Revision |
Given the system of equations:
$$ 2 - 2 \lambda x = 0 $$ $$ 4 - 2 \lambda x = 0 $$ $$ x^2 + y^2 = 20 $$
We can simply solve for $x$ and $y$ in terms of $\lambda$ from the first two equations and plug it in to the third equation.
Solving for $x$: $$ 2 - 2 \lambda x = 0 $$ $$ 2 \lambda x = 2 $$ $$ x = \frac{1}{\lambda} $$
Solving for $y$: $$ 4 - 2 \lambda y = 0 $$ $$ 2 \lambda y = 4 $$ $$ y = \frac{2}{\lambda} $$
Plugging in to the third equation: $$ \left(\frac{1}{\lambda}\right)^2 + \left(\frac{2}{\lambda}\right)^2 = 20 $$ $$ \frac{1}{\lambda^2} + \frac{4}{\lambda^2} = 20 $$ $$ \frac{5}{\lambda^2} = 20 $$ $$ \lambda^2 = \frac{5}{20} = \frac{1}{4} $$ $$ \lambda = \pm \frac{1}{2} $$
This means that there are $2$ cases to examine. When $\lambda = -\frac{1}{2}$: $$ x = \frac{1}{-\frac{1}{2}} = -2 $$ $$ y = \frac{2}{-\frac{1}{2}} = -4 $$
When $\lambda = \frac{1}{2}$: $$ x = \frac{1}{\frac{1}{2}} = 2 $$ $$ y = \frac{2}{\frac{1}{2}} = 4 $$
The two points that solve this system of equations are $(-2, -4)$ and $(2, 4)$.
Comment: Awesome!!
