Question

Find the points of intersection of the conics x^2/4 - y^2/9 = 1 and x^2/6 + y^2/9 = 1.

Answer 1

Answer on Question #45090 – Math - Analytic Geometry

Problem.

Find the points of intersection of the conics $x^2/4 - y^2/9 = 1$ and $x^2/6 + y^2/9 = 1$.

Solution.

The points of intersection of the conics is the solution of the system

$$\left\{ \begin{array}{l} \frac {x ^ {2}}{4} - \frac {y ^ {2}}{9} = 1; \\ \frac {x ^ {2}}{6} + \frac {y ^ {2}}{9} = 1. \end{array} \right.$$

The system is equivalent to

$$\left\{ \begin{array}{l} \frac {x ^ {2}}{4} - \frac {y ^ {2}}{9} = 1; \\ \left(\frac {x ^ {2}}{6} + \frac {y ^ {2}}{9}\right) + \left(\frac {x ^ {2}}{4} - \frac {y ^ {2}}{9}\right) = 1 + 1; \end{array} \right.$$

or

$$\left\{ \begin{array}{l} \frac {x ^ {2}}{4} - \frac {y ^ {2}}{9} = 1; \\ \frac {x ^ {2}}{6} + \frac {x ^ {2}}{4} = 2; \end{array} \right.$$

Then $x^2 = 4.8$ and $y^2 = 1.8$.

Hence the points of intersection are $(- \sqrt{4.8}, - \sqrt{1.8})$, $(- \sqrt{4.8}, \sqrt{1.8})$, $(\sqrt{4.8}, - \sqrt{1.8})$, $(\sqrt{4.8}, \sqrt{1.8})$.

Answer: $(- \sqrt{4.8}, - \sqrt{1.8})$, $(- \sqrt{4.8}, \sqrt{1.8})$, $(\sqrt{4.8}, - \sqrt{1.8})$, $(\sqrt{4.8}, \sqrt{1.8})$.

www.AssignmentExpert.com