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Question
match the graphs to their equations. f
\\( \frac { ( x + 1 ) ^ { 2 } } { 16 } + \frac { ( y - 3 ) ^ { 2 } } { 1 } = 1 \\)
\\( \frac { ( x - 1 ) ^ { 2 } } { 16 } + \frac { ( y + 3 ) ^ { 2 } } { 1 } = 1 \\)
\\( \frac { ( x + 1 ) ^ { 2 } } { 1 } + \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1 \\)
\\( \frac { ( x - 1 ) ^ { 2 } } { 1 } + \frac { ( y - 3 ) ^ { 2 } } { 16 } = 1 \\)
To match the ellipse equations to their graphs, we use the standard form of an ellipse:
- Horizontal major axis: \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\) (\(a > b\)), center \((h, k)\), width \(2a\), height \(2b\).
- Vertical major axis: \(\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1\) (\(a > b\)), center \((h, k)\), width \(2b\), height \(2a\).
Equation 1: \(\frac{(x + 1)^2}{16} + \frac{(y - 3)^2}{1} = 1\)
- Center: \((h, k) = (-1, 3)\) (since \(x + 1 = x - (-1)\), \(y - 3\)).
- \(a^2 = 16 \implies a = 4\), \(b^2 = 1 \implies b = 1\).
- Horizontal major axis ( \(a > b\) ), width \(2a = 8\), height \(2b = 2\).
- Matches graph a (center near \((-1, 3)\), horizontal ellipse).
Equation 2: \(\frac{(x - 1)^2}{16} + \frac{(y + 3)^2}{1} = 1\)
- Center: \((h, k) = (1, -3)\) ( \(x - 1\), \(y + 3 = y - (-3)\) ).
- \(a = 4\), \(b = 1\) (horizontal major axis).
- Matches graph b (center near \((1, -3)\), horizontal ellipse).
Equation 3: \(\frac{(x + 1)^2}{1} + \frac{(y + 3)^2}{16} = 1\)
- Center: \((h, k) = (-1, -3)\) ( \(x + 1\), \(y + 3\) ).
- \(a^2 = 16 \implies a = 4\), \(b^2 = 1 \implies b = 1\).
- Vertical major axis ( \(a > b\) ), width \(2b = 2\), height \(2a = 8\).
- Matches graph with center \((-1, -3)\) and vertical stretch (not shown, but process holds).
Equation 4: \(\frac{(x - 1)^2}{1} + \frac{(y - 3)^2}{16} = 1\)
- Center: \((h, k) = (1, 3)\) ( \(x - 1\), \(y - 3\) ).
- \(a^2 = 16 \implies a = 4\), \(b^2 = 1 \implies b = 1\).
- Vertical major axis, width \(2b = 2\), height \(2a = 8\).
- Matches graph c (center near \((1, 3)\), vertical ellipse).
Graph Matching:
- \(\frac{(x + 1)^2}{16} + \frac{(y - 3)^2}{1} = 1\) → a
- \(\frac{(x - 1)^2}{16} + \frac{(y + 3)^2}{1} = 1\) → b
- \(\frac{(x + 1)^2}{1} + \frac{(y + 3)^2}{16} = 1\) → (vertical ellipse, center \((-1, -3)\))
- \(\frac{(x - 1)^2}{1} + \frac{(y - 3)^2}{16} = 1\) → c
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- \(\boldsymbol{\frac{(x + 1)^2}{16} + \frac{(y - 3)^2}{1} = 1}\) → a
- \(\boldsymbol{\frac{(x - 1)^2}{16} + \frac{(y + 3)^2}{1} = 1}\) → b
- \(\boldsymbol{\frac{(x + 1)^2}{1} + \frac{(y + 3)^2}{16} = 1}\) → (vertical ellipse, center \((-1, -3)\))
- \(\boldsymbol{\frac{(x - 1)^2}{1} + \frac{(y - 3)^2}{16} = 1}\) → c
(Note: For the third equation, if a third graph exists, it would have center \((-1, -3)\) and vertical stretch. The given graphs a, b, c suggest c is vertical with center \((1, 3)\), so the third equation likely matches an implicit graph with center \((-1, -3)\).)