Find the vertices and the equation of the conic section that has foci at (-5,0) and (5,0). It goes through the point (squareroot 18,4) and (squareroot 27, negative squareroot 32).
Thank you
Thank you
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Well with two foci, you immediately know it is either a hyperbola or a ellipse.
Now with the foci being equidistant from the origin, we know that the conic must have the standard form
x²/a² + y²/b² = 1 for ellipse
or
x²/a² - y²/b² = 1 for hyperbola
Note that the reason I know it is not y²/b² - x²/a² = 1 is because the foci lies on the x-axis and it will always lie on the focal axis.
Now we could just plug in the numbers to see which one fits, but if you actually plot the points, you will see that it cannot be an ellipse.
So it must be the hyperbola.
Plug in the points
(1) (3√2)²/a² - (-4)²/b² = 1
(2) (3√3)²/a² - (-4√2)²/b² = 1
Solving you get b² = (4/3)²a²
Put this into eqtn (1) and you should find that a = 3 and b = 4
So the equation is x²/3² - y²/4² = 1 and the vertices are (±3,0)
Yin
Now with the foci being equidistant from the origin, we know that the conic must have the standard form
x²/a² + y²/b² = 1 for ellipse
or
x²/a² - y²/b² = 1 for hyperbola
Note that the reason I know it is not y²/b² - x²/a² = 1 is because the foci lies on the x-axis and it will always lie on the focal axis.
Now we could just plug in the numbers to see which one fits, but if you actually plot the points, you will see that it cannot be an ellipse.
So it must be the hyperbola.
Plug in the points
(1) (3√2)²/a² - (-4)²/b² = 1
(2) (3√3)²/a² - (-4√2)²/b² = 1
Solving you get b² = (4/3)²a²
Put this into eqtn (1) and you should find that a = 3 and b = 4
So the equation is x²/3² - y²/4² = 1 and the vertices are (±3,0)
Yin
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