The latus rectum of parabola $y^{2} = 5 x + 4 y + 1$ is

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Solution

$y^{2} - 4 y + 4 = 5 x + 5 \Rightarrow (y - 2)^{2} = 5 (x + 1)$

Obviously, latus rectum is 5.

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$\\(i)~y^2=8x\\(ii)~4x^2+y=0\\(iii)~y^2-4y-3x+1=0\\(iv)~y^2-4y+4x=0\\(v)~~y^2+4x+4y-3=0\\(vi)~y^2=8x+8y\\(vii)~4(y-1)^2=-7(x-3)\\(viii)~y^2=5x-4y-9\\(ix)~x^2+y=6x-14$

The latus rectum of parabola $y^{2} = 5 x + 4 y + 1$ is

Latus rectum of the parabola $y^{2} - 4 y - 2 x - 8 = 0$ is

Q. The length of the latus rectum of the parabola

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is:

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The latus rectum of parabola y2=5x+4y+1 is

y2−4y+4=5x+5⇒(y−2)2=5(x+1) Obviously, latus rectum is 5.

The latus rectum of parabola $y^{2} = 5 x + 4 y + 1$ is

$y^{2} - 4 y + 4 = 5 x + 5 \Rightarrow (y - 2)^{2} = 5 (x + 1)$

Obviously, latus rectum is 5.