Question

Plot the following curves : (i) \(y=\pm \sqrt{x^{2}-1}\) (ii) \(y=2 \pm \sqrt{(x-1)^{2}-1}\) (iii) \(\mathrm{y}=\pm \sqrt{\mathrm{x}^{2}+1}\) (iv) \(4 y^{2}+4 y-x^{2}=0\)

Short answer

Answer: The given curve \(4y^2 + 4y - x^2 = 0\) represents an ellipse with a vertical major axis of length \(2b = \frac{2}{\sqrt{2}}\) and a horizontal minor axis length of \(2a = 2\). The center of the ellipse is located at the point \((0, -\frac{1}{2})\).

Step-by-step solution

Curve (i): \(y=\pm \sqrt{x^{2}-1}\)

To plot the curve, let's first analyze the function \(y=\pm \sqrt{x^{2}-1}\). 1. The domain of the function: Since the square root of a negative number is undefined, the domain will be \(x^{2}-1 \geq 0 \Rightarrow x^{2} \geq 1\). So, the domain is \(x \geq 1\) or \(x \leq -1\). 2. There are two separate function graphs based on the positive and negative square roots. Now, we can sketch the graph: 1. Plot points where \(x=1, -1\). When \(x=1\), \(y=\pm 0\). When \(x=-1\), \(y=\pm 0\). 2. As \(x\) gets larger, the value of \(y\) increases for the positive square root and decreases for the negative square root. 3. Plot the graph accordingly with two separate curves for the positive and negative square roots.

Curve (ii): \(y=2 \pm \sqrt{(x-1)^{2}-1}\)

To plot the curve, we will first analyze the given function. 1. The domain of this function: Since the square root of a negative number is undefined, the domain will be \((x-1)^{2}-1 \geq 0 \Rightarrow (x-1)^{2} \geq 1\). So, the domain is \(x \geq 2\) or \(x \leq 0\). 2. There are two separate function graphs based on the positive and negative square roots. Now, we can sketch the graph: 1. Plot points where \(x=2, 0\). When \(x=2\), \(y=3, 1\). When \(x=0\), \(y=3, 1\). 2. As \(x\) gets larger, the value of \(y\) increases for the positive square root and decreases for the negative square root. 3. Plot the graph accordingly with two separate curves for the positive and negative square roots.

Curve (iii): \(\mathrm{y}=\pm \sqrt{\mathrm{x}^{2}+1}\)

To plot this curve, let's first analyze the given function. 1. The domain of this function: Since the square root of a negative number is undefined, the value under the square root is always positive (\(x^2 + 1\)). Therefore, the domain will be all real numbers. 2. There are two separate function graphs based on the positive and negative square roots. Now, we can sketch the graph: 1. Plot the points where \(x=0\). When \(x=0\), \(y=\pm 1\). 2. As \(x\) gets larger, the value of \(y\) increases for the positive square root and decreases for the negative square root. 3. Plot the graph accordingly with two separate curves for the positive and negative square roots.

Curve (iv): \(4 y^{2}+4 y-x^{2}=0\)

To plot the curve, we can rewrite the equation: 1. Add \(x^2\) and 1 to both sides: \(x^2 + 4y^2 + 4y + 1 = 1\). 2. Factor the left side of the equation: \((x^2 + 1) + 4(y^2 + y + \frac{1}{4}) = 1 + 4(\frac{1}{4})\). 3. Complete the square for the \(y\) terms: \((x^2 + 1) + 4(y + \frac{1}{2})^2 = 2\). 4. Divide both sides by 2: \(\frac{(x^2+1)}{1} + \frac{4(y + \frac{1}{2})^2}{2} = 1\). Now this equation is in the form of a standard ellipse equation: \(\frac{x^2}{a^2} + \frac{(y - h)^2}{b^2} = 1\). From there, we can see that \(a=1\), \(b=\frac{1}{\sqrt{2}}\), and \(h=-\frac{1}{2}\). So, the ellipse has a vertical major axis of length \(2b=2/\sqrt{2}\) and a horizontal minor axis length of 2a=2 centered at \((0,-\frac{1}{2})\). Now, we can sketch the graph: 1. Identify the center of the ellipse: \((0,-\frac{1}{2})\). 2. Draw the major and minor axes with their corresponding lengths (2 and \(\frac{2}{\sqrt{2}}\)). 3. Plot the ellipse surrounding the center with the given axis lengths.