4. An ellipse has foci at $$ F_{1}(-1,0) $$ and $$ F_{2}(0, \sqrt{3}) $$. Let $$ q $$ be the distance from the centre to the vertices. (a) Write down the general equation for the conic section in terms of $$ q $$. (b) If the constant term of the equation in part (i) is -33 , find the value of $$ q $$. (c) In the rotated $$ X Y $$-coordinate system, one vertex of the conic section is at $$ \left(\frac{5}{2}, \frac{\sqrt{3}}{2}\right) $$. Find the corresponding coordinates of this vertex in the $$ x y $$-coordinate system.

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UpStudy AI · Accepted Answer

To solve the problem step by step, let's start with the known conditions:

1. The foci of the ellipse are at $$ F_{1}(-1,0) $$ and $$ F_{2}(0, \sqrt{3}) $$.
2. The distance from the center to the vertices is denoted as $$ q $$.
3. We need to find the general equation of the ellipse, the value of $$ q $$ when the constant term is -33, and the coordinates of a vertex in the $$ xy $$-coordinate system.

### Part (a): General Equation for the Conic Section

The general equation of an ellipse centered at the origin with foci at $$ (c, 0) $$ and $$ (-c, 0) $$ is given by:

$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
$$

where:
- $$ c $$ is the distance from the center to each focus,
- $$ a $$ is the distance from the center to each vertex along the major axis,
- $$ b $$ is the distance from the center to each vertex along the minor axis.

For an ellipse, the relationship between $$ a $$, $$ b $$, and $$ c $$ is given by:

$$
c^2 = a^2 - b^2
$$

Given that $$ q $$ is the distance from the center to the vertices, we can set $$ a = q $$.

Next, we need to find $$ c $$ based on the foci. The distance between the foci can be calculated as:

$$
c = \sqrt{(-1 - 0)^2 + (0 - \sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
$$

Thus, we have $$ c = 2 $$.

Now, substituting $$ a = q $$ into the relationship $$ c^2 = a^2 - b^2 $$:

$$
2^2 = q^2 - b^2 \implies 4 = q^2 - b^2 \implies b^2 = q^2 - 4
$$

The general equation of the ellipse becomes:

$$
\frac{x^2}{q^2} + \frac{y^2}{q^2 - 4} = 1
$$

### Part (b): Finding the Value of $$ q $$

We are given that the constant term of the equation is -33. The standard form of the ellipse can be rearranged to:

$$
\frac{x^2}{q^2} + \frac{y^2}{q^2 - 4} = 1 \implies -33 = -\left(q^2 - 4\right) \implies q^2 - 4 = 33
$$

Solving for $$ q^2 $$:

$$
q^2 = 33 + 4 = 37
$$

Thus, $$ q = \sqrt{37} $$.

### Part (c): Finding the Corresponding Coordinates in the $$ xy $$-Coordinate System

To find the coordinates of the vertex $$ \left(\frac{5}{2}, \frac{\sqrt{3}}{2}\right) $$ in the rotated $$ XY $$-coordinate system, we need to determine the rotation angle. The foci suggest a rotation, and we can find the angle $$ \theta $$ using the coordinates of the foci.

The angle $$ \theta $$ can be calculated as:

$$
\tan(\theta) = \frac{\sqrt{3}}{-1} \implies \theta = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}
$$

To convert the coordinates back to the $$ xy $$-coordinate system, we can use the rotation matrix:

$$
\begin{pmatrix}
x \
y
\end{pmatrix}
=
\begin{pmatrix}
\cos(\theta) & -\sin(\theta) \
\sin(\theta) & \cos(\theta)
\end{pmatrix}
\begin{pmatrix}
X \
Y
\end{pmatrix}
$$

Substituting $$ \theta = -\frac{\pi}{3} $$:

$$
\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}
$$

Thus, the rotation matrix becomes:

$$
\begin{pmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} \
-\frac{\sqrt{3}}{2} & \frac{1}{2}
\end{pmatrix}
$$

Now, applying this to the vertex $$ \left(\frac{5}{2}, \frac{\sqrt{3}}{2}\right) $$:

$$
\begin{pmatrix}
x \
y
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} \
-\frac{\sqrt{3}}{2} & \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
\frac{5}{2} \
\frac{\sqrt{3}}{2}
\end{pmatrix}
$$

Calculating the components:

1. For $$ x $$:

$$
x = \frac{1}{2} \cdot \frac{5}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{5}{4} + \frac{3}{4} = \frac{8}{4} = 2
$$

2. For $$ y $$:

$$
y = -\frac{\sqrt{3}}{2} \cdot \frac{5}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{5\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = -\frac{4\sqrt{3}}{4} = -\sqrt{3}
$$

Thus, the corresponding coordinates of the vertex in the $$ xy $$-coordinate system are $$ (2, -\sqrt{3}) $$.

### Summary of Results
(a) The general equation is $$ \frac{x^2}{q^2} + \frac{y^2}{q^2 - 4} = 1 $$.

(b) The value of $$ q $$ is $$ \sqrt{37} $$.

(c) The coordinates of the vertex in the $$ xy $$-coordinate system are $$ (2, -\sqrt{3}) $$.
(a) The general equation of the ellipse is $$ \frac{x^2}{q^2} + \frac{y^2}{q^2 - 4} = 1 $$. (b) The value of $$ q $$ is $$ \sqrt{37} $$. (c) The coordinates of the vertex in the $$ xy $$-coordinate system are $$ (2, -\sqrt{3}) $$.

Upstudy AI Solution

Answer

Solution

Part (a): General Equation for the Conic Section

Part (b): Finding the Value of

Part ©: Finding the Corresponding Coordinates in the -Coordinate System

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Summary of Results

Beyond the Answer

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