Solving Quadratic System: 4x^2-y^2-8x+6y-9=0, 2x^2-3y^2+4x+

[cheyla13]

These are the directions: Find the points of intersection, if any, of the graphs in the system.

The system(s) is (are):
4x[sup:3vncvica]2[/sup:3vncvica]-y[sup:3vncvica]2[/sup:3vncvica]-8x+6y-9=0
2x[sup:3vncvica]2[/sup:3vncvica]-3y[sup:3vncvica]2[/sup:3vncvica]+4x+18y-43=0

Using Linear Combinations, I got: 2x[sup:3vncvica]2[/sup:3vncvica]+2y[sup:3vncvica]2[/sup:3vncvica]-12x-12y+34
I don't think this is the simplest form...
Help.... :?

 

[royhaas]

The points of intersection will lie on the circle. Determine its equation first.

 

[soroban]

Hello, cheyla13!

This is a tricky one . . .

I'll solve it completely ... and hope this helps you solve the others.

Find the points of intersection, if any.

. . \(\displaystyle \begin{array}{cccc}4x^2 - y^2 -8x+6y-9 &=& 0 & {\bf[1]} \\ 2x^2-3y^2+4x+18y-43&=&0 & {\bf[2]}\end{array}\)

\(\displaystyle \text{{\bf[1]} becomes: }\;4(x^2 - 2x\quad\;\ - (y^2 - 6y\quad\;\ \:=\:9\)

. . . . . . . . .\(\displaystyle 4(x^2 - 2x + 1) - (y^2 - 6y + 9) \:=\:9 + 4 - 9\)

. . . . . . . . . . . . . . . .\(\displaystyle 4(x-1)^2 - (y-3)^2 \:=\:4\;\;{\bf[3]}\)


\(\displaystyle \text{{\bf[2]} becomes: }\;2(x^2 + 2x\quad\;\ - 3(y^2-6y\quad\;\ \:=\:43\)

. . . . . . . . .\(\displaystyle 2(x^2 + 2x + 1) - 3(y^2 - 6y + 9) \:=\:43 + 2 - 27\)

. . . . . . . . . . . . . . . .\(\displaystyle 2(x + 1)^2 - 3(y-3)^2\:=\:18\;\;{\bf[4]}\)


\(\displaystyle \begin{array}{cccc}\text{Multiply {\bf[3]} by 3:} & 12(x-1)^2 - 3(y-3)^2 &=& 12 \\ \text{Subtract {\bf[4]}:} & 2(x+1)^2 - 3(y-3)^2 &=& 18 \end{array}\)

\(\displaystyle \text{And we have: }\;12(x-1)^2 - 2(x+1)^2 \:=\:-6\)

. . \(\displaystyle \text{which simplifies to: }\;5x^2 - 14x + 8 \:=\:0\)

. . \(\displaystyle \text{which factors: }\;(x-2)(5x-4) \:=\:0\)

. . \(\displaystyle \text{and has roots: }\;x \;=\;2,\:\frac{4}{5}\)


\(\displaystyle \text{Substitute }x=2\text{ into {\bf[1]}: }\; 16-y^2-16+6y-9\:=\:0 \quad\Rightarrow\quad y^2-6y + 9 \:=\:0\)

. . . \(\displaystyle (y-3)^2 \:=\:0\quad\Rightarrow\quad y \:=\:3\)

\(\displaystyle \text{One point of intersection is: }\;(2,\,3)\)


\(\displaystyle \text{Substitute }x=\frac{4}{5}\text{ into {\bf[1]}: }\;4\left(\frac{16}{25}\right) - y^2 - 8\left(\frac{4}{5}\right) + 6y - 9 \:=\:0\)

. . \(\displaystyle \text{which simplifies to: }\;25y^2 - 150y + 321 \:=\:0\quad\hdots \text{ which has no real roots.}\)


\(\displaystyle \text{Therefore, the only intersection is:} \;(2,\,3)\)

 

[Denis]

4x^2 - y^2 - 8x + 6y - 9 = 0 [1]
2x^2 -3y^2 + 4x + 18y - 43 = 0 [2]


Or multiply [1] by 3:
12x^2 - 3y^2 - 24x + 18y - 27 = 0 [1]
2x^2 - 3y^2 + 4x + 18y - 43 = 0 [2]

Subtract 'em: 10x^2 - 28x + 16 = 0
Simplify: 5x^2 - 14x + 8 = 0
Factor: (x - 2)(5x - 4) = 0 : x = 2 or 4/5 ; go back and get y