Question

sketch a graph of $\frac{(x + 1)^2}{1}+\frac{(y + 3)^2}{4}=1$

Explanation:

Step1: Identify the standard form of ellipse

The standard form of an ellipse is \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\) (\(a>b>0\)), where \((h,k)\) is the center of the ellipse. For the given equation \(\frac{(x + 1)^2}{1}+\frac{(y+3)^2}{4}=1\), we have \(h=-1\), \(k = - 3\), \(a = 2\), \(b = 1\).

Step2: Find the center

The center of the ellipse \((h,k)=(-1,-3)\).

Step3: Find the vertices

The vertices of the ellipse (since the major - axis is along the \(y\) - axis as \(a>b\)) are \((h,k\pm a)\). Substituting the values, we get \((-1,-3 + 2)=(-1,-1)\) and \((-1,-3-2)=(-1,-5)\).

Step4: Find the co - vertices

The co - vertices of the ellipse are \((h\pm b,k)\). Substituting the values, we get \((-1 + 1,-3)=(0,-3)\) and \((-1-1,-3)=(-2,-3)\).

Answer:

Plot the center at \((-1,-3)\), vertices at \((-1,-1)\) and \((-1,-5)\), co - vertices at \((0,-3)\) and \((-2,-3)\) and then draw the ellipse passing through these points.