Question

(a^2+b^2)\ab+1=4
Give a equation to solve the value of a and b.
a=? & b?

Answer 1

Answer on Question #83959 - Math - Analytic Geometry

Question

$$\frac {a ^ {2} + b ^ {2}}{a b + 1} = 4$$

Give a equation to solve the value of $a$ and $b$ .

$a = ?$ & $b?$

Solution

$$a ^ {2} + b ^ {2} - 4 a b - 4 = 0$$ $$a _ {1 1} a ^ {2} + 2 a _ {1 2} a b + 2 a _ {1 3} a + a _ {2 2} b ^ {2} + 2 a _ {2 3} b + a _ {3 3} = 0$$ $$a _ {1 1} = 1$$ $$a _ {1 2} = - 2$$ $$a _ {1 3} = 0$$ $$a _ {2 2} = 1$$ $$a _ {2 3} = 0$$ $$a _ {3 3} = - 4$$

Then:

$$I _ {1} = a _ {1 1} + a _ {2 2} = 1 + 1 = 2$$ $$I _ {2} = \left| \begin{array}{c c} a _ {1 1} & a _ {1 2} \\ a _ {1 2} & a _ {2 2} \end{array} \right| = \left| \begin{array}{c c} 1 & - 2 \\ - 2 & 1 \end{array} \right| = 1 - 4 = - 3$$ $$I _ {3} = \left| \begin{array}{c c c} a _ {1 1} & a _ {1 2} & a _ {1 3} \\ a _ {1 2} & a _ {2 2} & a _ {2 3} \\ a _ {1 3} & a _ {2 3} & a _ {3 3} \end{array} \right| = \left| \begin{array}{c c c} 1 & - 2 & 0 \\ - 2 & 1 & 0 \\ 0 & 0 & - 4 \end{array} \right| = - 4 + 2 \cdot 8 = 1 2$$

Since $I_2 < 0$ and $I_3 \neq 0$ , then we have equation of hyperbola.

So:

$$I (\lambda) = \left| \begin{array}{c c} - \lambda + 1 & - 2 \\ - 2 & - \lambda + 1 \end{array} \right| = (- \lambda + 1) ^ {2} - 4 = \lambda^ {2} - 2 \lambda + 1 - 4 = \lambda^ {2} - 2 \lambda - 3 = 0$$ $$\lambda_ {1} = 3$$ $$\lambda_ {2} = - 1$$

For $\lambda_{1} = 3$ one gets $a_2 = -a_1$ , hence $v_{1} = \binom{1}{-1}$ , $e_1 = \frac{v_1}{\|v_1\|} = \frac{1}{\sqrt{1^2 + (-1)^2}}\binom{1}{-1} = \binom{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}$

For $\lambda_{2} = -1$ one gets $a_2 = a_1$ , hence $v_{2} = \binom{1}{1}$ , $e_2 = \frac{v_2}{\|v_2\|} = \frac{1}{\sqrt{1^2 + 1^2}}\binom{1}{1} = \binom{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}$

Let $R = \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right)$ . Then the rotation is $\binom{a}{b} = R \left( \begin{array}{c} \tilde{a} \\ \tilde{b} \end{array} \right) = \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right) \left( \begin{array}{c} \tilde{a} \\ \tilde{b} \end{array} \right) = \left( \begin{array}{c} \frac{\tilde{a}}{\sqrt{2}} + \frac{\tilde{b}}{\sqrt{2}} \\ -\frac{\tilde{a}}{\sqrt{2}} + \frac{\tilde{b}}{\sqrt{2}} \end{array} \right)$ ,

hence $\left( \begin{array}{c}\tilde{a}\\ \tilde{b} \end{array} \right) = R^{-1}\left( \begin{array}{c}a\\ b \end{array} \right) = \left( \begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right)^{-1}\left( \begin{array}{c}a\\ b \end{array} \right) = \left( \begin{array}{cc}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right)\left( \begin{array}{c}a\\ b \end{array} \right) = \left( \begin{array}{c}\frac{a}{\sqrt{2}} -\frac{b}{\sqrt{2}}\\ \frac{a}{\sqrt{2}} +\frac{b}{\sqrt{2}} \end{array} \right)$

Substituting $a = \frac{\tilde{a}}{\sqrt{2}} + \frac{\tilde{b}}{\sqrt{2}}$ , $b = -\frac{\tilde{a}}{\sqrt{2}} + \frac{\tilde{b}}{\sqrt{2}}$ into the equation $a^2 + b^2 - 4ab - 4 = 0$ one gets

$$\left(\frac {\tilde {a}}{\sqrt {2}} + \frac {\tilde {b}}{\sqrt {2}}\right) ^ {2} + \left(- \frac {\tilde {a}}{\sqrt {2}} + \frac {\tilde {b}}{\sqrt {2}}\right) ^ {2} - 4 \left(\frac {\tilde {a}}{\sqrt {2}} + \frac {\tilde {b}}{\sqrt {2}}\right) \left(- \frac {\tilde {a}}{\sqrt {2}} + \frac {\tilde {b}}{\sqrt {2}}\right) - 4 = 0$$ $$\frac {1}{2} (\tilde {a}) ^ {2} + \tilde {a} \tilde {b} + \frac {1}{2} (\tilde {b}) ^ {2} + \frac {1}{2} (\tilde {a}) ^ {2} - \tilde {a} \tilde {b} + \frac {1}{2} (\tilde {b}) ^ {2} - 2 \left(\left(\tilde {b}\right) ^ {2} - (\tilde {a}) ^ {2}\right) - 4 = 0$$ $$3 \tilde {a} ^ {2} - \tilde {b} ^ {2} - 4 = 0$$

One gets an equation of hyperbola:

$$\frac {\tilde {a} ^ {2}}{\frac {4}{3}} - \frac {\tilde {b} ^ {2}}{4} = 1$$

The canonical form of the equation:

$$\tilde {a} ^ {2} \lambda_ {1} + \tilde {b} ^ {2} \lambda_ {2} + \frac {l _ {3}}{l _ {2}} = 0$$ $$3 \tilde {a} ^ {2} - \tilde {b} ^ {2} - 4 = 0$$ $$\frac {\tilde {a} ^ {2}}{\frac {4}{3}} - \frac {\tilde {b} ^ {2}}{4} = 1$$

Answer: $\frac{\tilde{a}^2}{\frac{4}{3}} - \frac{\tilde{b}^2}{4} = 1$

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