The length of the latus rectum of the parabol | vedclass

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Question

(C) Given equation: $4y^2 - 20y + 2x + 17 = 0$ Divide by $4$: $y^2 - 5y + \frac{1}{2}x + \frac{17}{4} = 0$ Complete the square for $y$: $(y - \frac{5}

Vedclass · Accepted Answer

$0.5$

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(C) Given equation: $4y^2 - 20y + 2x + 17 = 0$ Divide by $4$: $y^2 - 5y + \frac{1}{2}x + \frac{17}{4} = 0$ Complete the square for $y$: $(y - \frac{5}{2})^2 - \frac{25}{4} + \frac{1}{2}x + \frac{17}{4} = 0$ $(y - \frac{5}{2})^2 = -\frac{1}{2}x + 2$ $(y - \frac{5}{2})^2 = -\frac{1}{2}(x - 4)$ Comparing with $(y - k)^2 = -4a(x - h)$,we get $4a = \frac{1}{2} = 0.5$. Thus,the length of the latus rectum is $0.5$.

The length of the latus rectum of the parabola $4y^2 + 2x - 20y + 17 = 0$ is

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