What is a^2-5a+3b-7=0 ?

The solution to a^2-5a+3b-7=0 is Parabola with (h,k)=(5/2 , 53/12),p=-3/4

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a^2-5a+3b-7=0

a^{2} - 5 a + 3 b - 7 = 0

(h, k) = (\frac{5}{2}, \frac{53}{12}), p = - \frac{3}{4}

Solution steps

a^{2} - 5 a + 3 b - 7 = 0

Rewrite

a^{2} - 5 a + 3 b - 7 = 0

in the standard form

4 (- \frac{3}{4}) (b - \frac{53}{12}) = (a - \frac{5}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{5}{2}, \frac{53}{12}), p = - \frac{3}{4}

16 x + 4 y^{2} - 90 = 0

(x - 1)^{2} + (y - 2) = 20

x^{2} + 8 x - 6 y + 28 = 0

y^{2} = 50 x

- x^{2} - 3 x + 5 = - x + k (x)

What is a^2-5a+3b-7=0 ?
The solution to a^2-5a+3b-7=0 is Parabola with (h,k)=(5/2 , 53/12),p=-3/4