# pre cal

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find the standard form of the equation of a parabola with the vertex (0,0) and the directrix y=6

find the standard form of the equation of the ellipse with center (0,0) and foci at (+- 4square root 3, 0) and verticies (+-8,0)

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a. An ellipse is formed by stretching the graph of x^2+ y^2=1 horizontally by a factor of 3 and vertically by a factor of 4. Determine the equation of the ellipse in standard form x^2/a^2+y^2/b^2=1 (x/3) ^2+ (y/4) ^2=1 x^2/9 + y^2/16 …
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a. An ellipse is formed by stretching the graph of x^2+ y^2=1 horizontally by a factor of 3 and vertically by a factor of 4. Determine the equation of the ellipse in standard form x^2/a^2+y^2/b^2=1 (x/3) ^2+ (y/4) ^2=1 x^2/9 + y^2/16 …
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find the standard form of the equation of the ellipse with center (0,0) and foci at (+- 4square root 3, 0) and verticies (+-8,0) x^2/a^2 + =y^2/b^2 = 1 a = 8 b = sqrt(8^2 - (4sqrt3)^2) ..= sqrt(64 - 48) = sqrt(16) = 4 Thus, x^2/64 …
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