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

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Question

(A) The equation of the given parabola is $2y^2 + 5x - 6y + 1 = 0$. Rearranging the terms,we get $2(y^2 - 3y) = -5x - 1$. Completing the square for $y

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

Vedclass · Answer

(A) The equation of the given parabola is $2y^2 + 5x - 6y + 1 = 0$. Rearranging the terms,we get $2(y^2 - 3y) = -5x - 1$. Completing the square for $y$: $2(y^2 - 3y + \frac{9}{4}) = -5x - 1 + \frac{9}{2}$. $2(y - \frac{3}{2})^2 = -5x + \frac{7}{2}$. $2(y - \frac{3}{2})^2 = -5(x - \frac{7}{10})$. $(y - \frac{3}{2})^2 = 4(-\frac{5}{8})(x - \frac{7}{10})$. Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,we have the vertex $(h, k) = (\frac{7}{10}, \frac{3}{2})$ and $a = -\frac{5}{8}$. The focus is given by $(h + a, k) = (\frac{7}{10} - \frac{5}{8}, \frac{3}{2}) = (\frac{28 - 25}{40}, \frac{3}{2}) = (\frac{3}{40}, \frac{3}{2})$. Thus,the vertex and focus are $(\frac{7}{10}, \frac{3}{2})$ and $(\frac{3}{40}, \frac{3}{2})$ respectively.

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

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