Question

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

Explanation:

Step1: Identify the center

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

Step2: Determine \(a\) and \(b\)

From \(\frac{(x + 1)^{2}}{1}+\frac{(y + 3)^{2}}{4}=1\), we get \(a^{2}=4\), \(b^{2}=1\). Then \(a = 2\), \(b = 1\).

Step3: Find the vertices

Since the major axis is vertical (because \(a^{2}\) is under the \(y\) - term), the vertices are \((h,k\pm a)\). Substituting \(h=-1\), \(k=-3\), \(a = 2\), we get the vertices \((-1,-3 + 2)=(-1,-1)\) and \((-1,-3-2)=(-1,-5)\).

Step4: Find the co - vertices

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

Answer:

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