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Question
question
find the endpoints of the latus rectum of the parabola with equation $(x + 9)^2=-20(y + 5)$.
provide your answer below:
(□,□) and (□,□)
Step1: Identify the form of the parabola
The given equation \((x + 9)^2=-20(y + 5)\) is of the form \((x - h)^2 = 4p(y - k)\), where \((h,k)\) is the vertex of the parabola. Comparing, we have \(h=-9,k = - 5\) and \(4p=-20\), so \(p=-5\).
Step2: Find the vertex
The vertex of the parabola \((x - h)^2=4p(y - k)\) is \((h,k)\). So the vertex of the given parabola is \((-9,-5)\).
Step3: Determine the endpoints of the latus - rectum
For a parabola of the form \((x - h)^2 = 4p(y - k)\), the length of the latus - rectum is \(|4p|\) and the endpoints of the latus - rectum lie on the parabola at a distance of \(|p|\) from the focus in the horizontal direction. The focus of the parabola \((x - h)^2=4p(y - k)\) is \((h,k + p)\), here the focus is \((-9,-5-5)=(-9,-10)\).
The endpoints of the latus - rectum have the \(y\) - coordinate equal to the \(y\) - coordinate of the focus (\(y=-10\)).
Substitute \(y = - 10\) into the equation \((x + 9)^2=-20(y + 5)\):
\((x + 9)^2=-20(-10 + 5)\)
\((x + 9)^2=-20\times(-5)\)
\((x + 9)^2 = 100\)
\(x+9=\pm10\)
If \(x + 9 = 10\), then \(x=1\); if \(x + 9=-10\), then \(x=-19\).
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\((1,-10)\) and \((-19,-10)\)