Problem 55

Question

Exercises $53-56$ give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes. $$ y^{2}-x^{2}=1, \quad \text { left } 1, \text { down } 1 $$

Step-by-Step Solution

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Answer

New equation: $(y+1)^2 - (x+1)^2 = 1$. Center: $(-1, -1)$. Vertices: $(-1, 0), (-1, -2)$. Foci: $(-1, -1+\sqrt{2}), (-1, -1-\sqrt{2})$. Asymptotes: $y=x-2$, $y=-x$.

1Step 1: Understand the Original Equation

The given equation for the hyperbola is $y^2 - x^2 = 1$. This is a hyperbola in standard form centered at the origin $(0, 0)$ because it compares to the form $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$. Here, $a^2 = 1$ and $b^2 = 1$, and the center $(h, k)$ is originally at $(0, 0)$.

2Step 2: Apply the Shifts

Shift the hyperbola left by 1 unit and down by 1 unit from the original center $(0, 0)$:- Left 1 unit: The new center in the x-direction becomes $-1$.- Down 1 unit: The new center in the y-direction becomes $-1$.Thus, the new center is $(-1, -1)$.

3Step 3: Rewrite the Transformed Hyperbola Equation

With the new center $(-1, -1)$, adjust the original equation $y^2 - x^2 = 1$ to its transformed version. The general formula becomes $(y + 1)^2 - (x + 1)^2 = 1$.

4Step 4: Identify the Vertices

The vertices of the hyperbola originally $(0, \pm a)$ now become $(-1, -1 \pm 1)$. This gives the vertices at $(-1, 0)$ and $(-1, -2)$.

5Step 5: Calculate the Foci

For the original hyperbola, $c$ is calculated using $c^2 = a^2 + b^2$, so $c = \sqrt{1 + 1} = \sqrt{2}$. The foci, originally at $(0, \pm \sqrt{2})$, are shifted to $(-1, -1 \pm \sqrt{2})$, which gives them as $(-1, -1 + \sqrt{2})$ and $(-1, -1 - \sqrt{2})$.

6Step 6: Determine the Asymptotes

The original asymptotes are $y = \pm x$. Due to the shifts, the asymptotes become $y + 1 = \pm(x + 1)$, leading to the equations of the asymptotes: - $y = x$ gives $y = x$- $y = -x$ gives $y = -x$So, considering the center shift, the equations are really transformed to:- $y = x - 2$- $y = -x$.

7Step 7: Combine Results

The equation for the new hyperbola is $(y + 1)^2 - (x + 1)^2 = 1$. The new center is $(-1, -1)$, vertices are $(-1, 0)$ and $(-1, -2)$, foci are $(-1, -1 + \sqrt{2})$ and $(-1, -1 - \sqrt{2})$, and the asymptotes are $y = x - 2$ and $y = -x$.

Key Concepts

Equation of a hyperbolaHyperbola transformationHyperbola asymptotesHyperbola fociHyperbola vertices

Equation of a hyperbola

A hyperbola is a type of conic section defined by its equation. The typical standard form of a hyperbola is either

Horizontal hyperbola: $ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 $
Vertical hyperbola: $ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 $

Here, $(h, k)$ is the center of the hyperbola, $a^2$ and $b^2$ are the squared lengths of the semi-major and semi-minor axes of the hyperbola, respectively.
In the equation $y^2 - x^2 = 1$, we can see it's a vertical hyperbola centered at the origin, $(0, 0)$, with both axes lengths being 1 unit since $a^2 = b^2 = 1$. This results in a symmetrical shape around the axes.

Hyperbola transformation

A transformation involves shifting the hyperbola in the coordinate plane. The transformation we applied involved moving it left by 1 unit and down by 1 unit.
This can be done by adjusting the center $(h, k)$ in the equation template. Here’s how it works for the original hyperbola $y^2 - x^2 = 1$:

Left 1 unit: Subtract 1 from the x-component of the center.
Down 1 unit: Subtract 1 from the y-component of the center.

With these shifts, the new center becomes $(-1, -1)$. The transformed equation then follows the generic hyperbola expression, which turns into $(y+1)^2 - (x+1)^2 = 1$. This reflects the shape's adjusted position on the coordinate plane.

Hyperbola asymptotes

Asymptotes are lines that the hyperbola approaches but never touches. They guide the shape of a hyperbola.
For our original hyperbola, the asymptotes are given by lines through the origin, forming the equations $y = \pm x$. Since our transformation involves shifting the center to $(-1, -1)$, we need to adjust these lines.

Starting with $y = x$, shifting the line contents down by 1 and left by 1 yields the asymptote $y = x - 2$.
For $y = -x$, shifting gives $y = -x$.

These updated equations represent the asymptotes of the hyperbola in its new position.

Hyperbola foci

The foci are special points located along the axis of symmetry of a hyperbola that play a crucial role in its geometric definition.
To determine the foci of our original hyperbola, we use the relationship $c^2 = a^2 + b^2$. Given that $a^2 = 1$ and $b^2 = 1$ for the original hyperbola, we find:

$c = \sqrt{1 + 1} = \sqrt{2}$

The original foci at $(0, \pm \sqrt{2})$ are transformed by shifting:

Add $-1$ to both the x and y coordinates.

So, the new foci become $(-1, -1 + \sqrt{2})$ and $(-1, -1 - \sqrt{2})$. These points are critical as they maintain the properties of the hyperbola even in its transformed state.

Hyperbola vertices

Vertices are points on the hyperbola that lie at the intersection of its major axis with the hyperbola.
For a vertical hyperbola like $y^2 - x^2 = 1$, the vertices originally at $(0, \pm 1)$ imply that $a = 1$, meaning the hyperbola extends 1 unit away from the center along the y-axis.