Answer:
A= –5x2 + 4x – 7y + 14 = 0
C= 9x2 – 12xy + 4y2 + 4x – y + 5 = 0
D= 12x + 6y – 5y2 + 14 = 0
Step-by-step explanation:
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The equation which would produce a parabola from the provided options are equation first and forth,
[tex]-5x^2 + 4x - 7y + 14 = 0[/tex]
[tex]12x + 6y -5y^2 + 14 = 0[/tex]
To of identify the type of conic section from an equation whether it is parabola, or not, let suppose the equation as,
[tex]Ax^2+By^2+Cx+Dy+E=0[/tex]
If In this equation,
Either [tex]A=0[/tex] or [tex]B=0[/tex], but not both. Then it is the equation of parabola.
The first equation is,
[tex]-5x^2 + 4x - 7y + 14 = 0[/tex]
There is no square term of y. Thus, this is the equation of parabola.
The second equation is,
[tex]-x^2 + 6x + y^2 - 8 = 0[/tex]
This is not the equation of parabola, as there is two square terms of variable x and y.
The third equation is,
[tex]9x^2 - 12xy + 4y^2 + 4x - y + 5 = 0[/tex]
Similar to the second equation, here is also two square terms of different variable. Thus, it is not a equation of parabola.
The fourth equation is,
[tex]12x + 6y -5y^2 + 14 = 0[/tex]
This equation is the equation of parabola, as there is only a single square term.
The fifth equation is,
[tex]20x^2 + 10xy - y^2 + 3x - 6y + 5 = 0[/tex]
This is now the equation of parabola because of two square terms.
Hence, the equation which would produce a parabola from the provided options are equation first and forth,
[tex]-5x^2 + 4x - 7y + 14 = 0[/tex]
[tex]12x + 6y -5y^2 + 14 = 0[/tex]
Learn more about the conic section here;
https://brainly.com/question/8320698
