The vertex and the focus of the parabola 2x^2 | vedclass

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(D) Given equation: $2x^2 + 5y - 6x + 1 = 0$ Divide by $2$: $x^2 - 3x + \frac{5}{2}y + \frac{1}{2} = 0$ Rearrange terms: $x^2 - 3x = -\frac{5}{2}y - \

Vedclass · Accepted Answer

$\left(\frac{3}{2}, \frac{7}{10}\right), \left(\frac{3}{2}, \frac{3}{40}\right)$

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(D) Given equation: $2x^2 + 5y - 6x + 1 = 0$ Divide by $2$: $x^2 - 3x + \frac{5}{2}y + \frac{1}{2} = 0$ Rearrange terms: $x^2 - 3x = -\frac{5}{2}y - \frac{1}{2}$ Complete the square: $x^2 - 3x + (\frac{3}{2})^2 = -\frac{5}{2}y - \frac{1}{2} + \frac{9}{4}$ $(x - \frac{3}{2})^2 = -\frac{5}{2}y + \frac{7}{4}$ $(x - \frac{3}{2})^2 = -\frac{5}{2}(y - \frac{7}{10})$ Comparing with $(x - h)^2 = -4a(y - k)$,we get $h = \frac{3}{2}$,$k = \frac{7}{10}$,and $4a = \frac{5}{2} \Rightarrow a = \frac{5}{8}$. The vertex is $(h, k) = (\frac{3}{2}, \frac{7}{10})$. The focus is $(h, k - a) = (\frac{3}{2}, \frac{7}{10} - \frac{5}{8}) = (\frac{3}{2}, \frac{28 - 25}{40}) = (\frac{3}{2}, \frac{3}{40})$. Thus,the vertex and focus are $(\frac{3}{2}, \frac{7}{10})$ and $(\frac{3}{2}, \frac{3}{40})$.

The vertex and the focus of the parabola $2x^2 + 5y - 6x + 1 = 0$ respectively,are

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