Step 1: Standardizing the Hyperbola Equation

Given the equation: $9y^2 - 4x^2 = 36$

To convert this into the standard form of a hyperbola, divide both sides by $36$ [Per the algebraic axiom of equivalent division]:

$\frac{9y^2}{36} - \frac{4x^2}{36} = \frac{36}{36}$

$\frac{y^2}{4} - \frac{x^2}{9} = 1$

Comparing this to the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we observe that the transverse axis is along the $y$-axis [Geometric property of vertical hyperbolas].

Step 2: Identifying Parameters $a$ and $b$

From the standard equation, we identify:

• $a^2 = 4 \implies a = 2$
• $b^2 = 9 \implies b = 3$

Step 3: Calculating Foci and Eccentricity

For a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, the relationship between parameters is defined by $c^2 = a^2 + b^2$ [Per the Pythagorean relation in hyperbolas].

$c^2 = 4 + 9 = 13 \implies c = \sqrt{13}$

Eccentricity ($e$):

$e = \frac{c}{a} = \frac{\sqrt{13}}{2}$

Foci Coordinates: $(0, \pm c) = (0, \pm \sqrt{13})$

Step 4: Determining Vertices and Latus Rectum

Vertices: $(0, \pm a) = (0, \pm 2)$

Length of Latus Rectum:

$L = \frac{2b^2}{a} = \frac{2(9)}{2} = 9$

Summary of Calculated Values

Final Solution: The foci are $(0, \pm \sqrt{13})$, the vertices are $(0, \pm 2)$, the eccentricity is $\frac{\sqrt{13}}{2}$, and the length of the latus rectum is $9$.

Step 1: Standardizing the Hyperbola Equation

Given the equation: $16x^2 - 9y^2 = 576$.

To convert this into the standard form of a hyperbola, $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we divide the entire equation by $576$:

$\frac{16x^2}{576} - \frac{9y^2}{576} = \frac{576}{576}$

$\frac{x^2}{36} - \frac{y^2}{64} = 1$ [By algebraic simplification]

Step 2: Identifying Parameters $a$ and $b$

Comparing the simplified equation with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

• $a^2 = 36 \implies a = 6$
• $b^2 = 64 \implies b = 8$

Step 3: Calculating Eccentricity ($e$) and Foci ($c$)

For a hyperbola, the relationship between $a, b,$ and $c$ is given by $c^2 = a^2 + b^2$ [Per the definition of hyperbola focal distance]:

$c^2 = 36 + 64 = 100$

$c = \sqrt{100} = 10$

The eccentricity ($e$) is defined as $e = \frac{c}{a}$:

$e = \frac{10}{6} = \frac{5}{3}$

Step 4: Deriving Geometric Coordinates and Measurements

Since the $x^2$ term is positive, the transverse axis lies along the x-axis.

• Vertices: $(\pm a, 0) = (\pm 6, 0)$
• Foci: $(\pm c, 0) = (\pm 10, 0)$
• Length of Latus Rectum: $LR = \frac{2b^2}{a}$ [Per standard conic formula]
• $LR = \frac{2(64)}{6} = \frac{128}{6} = \frac{64}{3}$

Summary of Results

Final Solution: The vertices are $(\pm 6, 0)$, the foci are $(\pm 10, 0)$, the eccentricity is $\frac{5}{3}$, and the length of the latus rectum is $\frac{64}{3}$ units.

Step 1: Analyzing Given Parameters

The vertices are given as $(\pm 6, 0)$ and the foci are given as $(\pm 4, 0)$.
Since the $y$-coordinates are zero for both vertices and foci, the major axis of the ellipse lies along the $x$-axis. [Per standard conic form criteria].

Step 2: Identifying Constants $a$ and $c$

The standard form for an ellipse centered at the origin with the major axis along the $x$-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$.

• The vertices are $(\pm a, 0)$. Given $(\pm 6, 0)$, we determine $a = 6$. Thus, $a^2 = 36$.
• The foci are $(\pm c, 0)$. Given $(\pm 4, 0)$, we determine $c = 4$. Thus, $c^2 = 16$.

Step 3: Calculating Parameter $b^2$

For any ellipse, the relationship between $a$, $b$, and $c$ is defined by the equation: $c^2 = a^2 - b^2$ [Derived from the focal property of ellipses].

Substituting the known values: $16 = 36 - b^2$ $b^2 = 36 - 16$ $b^2 = 20$

Step 4: Constructing the Equation

Substituting $a^2 = 36$ and $b^2 = 20$ into the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{36} + \frac{y^2}{20} = 1$

Step 5: Visual Representation

Final Solution: \frac{x^2}{36} + \frac{y^2}{20} = 1

Step 1: Establishing the Standard Form of a Circle

The standard equation of a circle with a center at $(h, k)$ and a radius $r$ is given by the formula:

[Per the definition of a circle as the locus of points at a constant distance from a fixed point].

Step 2: Identifying Given Parameters

From the problem statement, we identify the following variables:

• Center $(h, k) = (-a, -b)$
• Radius $r = \sqrt{a^2 - b^2}$

Step 3: Substitution into the General Equation

Substituting the values of $h$, $k$, and $r$ into the standard form $(x - h)^2 + (y - k)^2 = r^2$:

Step 4: Algebraic Simplification

Simplifying the signs and squaring the terms:

Expanding the binomial squares using the algebraic identity $(m + n)^2 = m^2 + 2mn + n^2$ [Per the Binomial Theorem/Algebraic Expansion]:

Step 5: Final Rearrangement

Subtract $a^2$ from both sides and add $b^2$ to both sides to consolidate the equation:

Visual Representation

Final Solution: x^2 + y^2 + 2ax + 2by + 2b^2 = 0

Step 1: Identification of the Parabolic Form

The given equation is $x^2 = 6y$. This equation follows the standard form of a parabola that opens upwards, expressed as $x^2 = 4ay$, where $a > 0$. [By the standard definition of a parabola symmetric about the $y$-axis].

Step 2: Determining the Value of the Parameter $a$

Comparing $x^2 = 6y$ with the standard equation $x^2 = 4ay$:

$4a = 6$

$a = \frac{6}{4} = 1.5$

[By algebraic comparison of coefficients].

Step 3: Calculating Geometric Properties

• Focus: For a parabola of the form $x^2 = 4ay$, the focus is located at $(0, a)$. Substituting $a = 1.5$, the focus is $(0, 1.5)$.
• Axis of the Parabola: Since the variable $x$ is squared, the parabola is symmetric about the $y$-axis. Thus, the axis of the parabola is the $y$-axis, represented by the equation $x = 0$.
• Equation of the Directrix: The directrix for this standard form is given by $y = -a$. Substituting $a = 1.5$, we get $y = -1.5$.
• Length of the Latus Rectum: The length of the latus rectum is defined as $|4a|$. Given $4a = 6$, the length is $6$ units.

Step 4: Geometric Representation

Step 5: Summary Table of Results

Final Solution: The coordinates of the focus are $(0, 1.5)$, the axis of the parabola is $x = 0$, the equation of the directrix is $y = -1.5$, and the length of the latus rectum is $6$ units.

Step 1: Analyzing Geometric Properties

Given the focus $F$ at $(6, 0)$ and the directrix $d$ at $x = -6$, we observe the following:

• The focus lies on the $x$-axis.
• The directrix is a vertical line perpendicular to the $x$-axis.
• Since the focus is $(a, 0)$ and the directrix is $x = -a$, the parabola is symmetric about the $x$-axis and opens towards the positive $x$-direction [Per the standard definition of a parabola].
• Here, $a = 6$.

Step 2: Mathematical Derivation via Conic Definition

By definition, a parabola is the locus of a point $P(x, y)$ such that its distance from the focus $F(6, 0)$ equals its perpendicular distance from the directrix $x = -6$.

Distance $PF = \sqrt{(x - 6)^2 + (y - 0)^2}$

Distance $PD = |x - (-6)| = |x + 6|$

Equating the distances (since $PF = PD$):

$\sqrt{(x - 6)^2 + y^2} = |x + 6|$

Squaring both sides [Axiom of Equality]:

$(x - 6)^2 + y^2 = (x + 6)^2$

Step 3: Algebraic Expansion and Simplification

Expanding the binomial squares:

$x^2 - 12x + 36 + y^2 = x^2 + 12x + 36$

Subtracting $x^2$ and $36$ from both sides:

$-12x + y^2 = 12x$

Adding $12x$ to both sides:

$y^2 = 12x + 12x$

$y^2 = 24x$

Visual Representation

Final Solution: The equation of the parabola is $y^2 = 24x$.

Step 1: Identifying the Standard Form

The given equation is $\frac{x^2}{36} + \frac{y^2}{16} = 1$. Comparing this with the standard equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$, we identify that this is a horizontal ellipse centered at the origin $(0,0)$.

Given parameters:

• $a^2 = 36 \implies a = 6$
• $b^2 = 16 \implies b = 4$

Step 2: Calculating Eccentricity ($e$)

The eccentricity of an ellipse is determined by the relationship $c^2 = a^2 - b^2$, where $c$ is the distance from the center to the foci.

$c^2 = 36 - 16 = 20$

$c = \sqrt{20} = 2\sqrt{5}$

The eccentricity $e$ is defined as $e = \frac{c}{a}$:

$e = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \approx 0.745$

Step 3: Calculating Key Geometric Properties

Step 4: Visualization

Final Solution: The foci are $(\pm 2\sqrt{5}, 0)$, the vertices are $(\pm 6, 0)$, the length of the major axis is $12$, the length of the minor axis is $8$, the eccentricity is $\frac{\sqrt{5}}{3}$, and the length of the latus rectum is $\frac{16}{3}$.

Step 1: Setting up the Coordinate System

We represent the parabolic cable on a Cartesian plane. Let the vertex of the parabola be at $(0, 6)$ on the $y$-axis. Since the bridge is $100\text{ m}$ long, the supporting wires extend from $x = -50$ to $x = 50$. The longest wires at the ends have a length of $30\text{ m}$, meaning the points $(\pm 50, 30)$ lie on the parabola [Definition of a Parabola with Vertical Axis].

Step 2: Defining the Parabolic Equation

The standard equation of a parabola with vertex at $(0, h)$ opening upwards is $x^2 = 4a(y - k)$. Given the vertex $(0, 6)$, the equation becomes:

$x^2 = 4a(y - 6)$

To find the constant $4a$, we substitute the point $(50, 30)$ which satisfies the equation [Method of Undetermined Coefficients]:

$50^2 = 4a(30 - 6)$

$2500 = 4a(24)$

$4a = \frac{2500}{24} = \frac{625}{6}$

Step 3: Calculating the Length at $x = 18$

The equation of the cable is $x^2 = \frac{625}{6}(y - 6)$. We need to find $y$ when $x = 18$ meters from the middle:

$18^2 = \frac{625}{6}(y - 6)$

$324 = \frac{625}{6}(y - 6)$

Isolating $(y - 6)$:

$y - 6 = \frac{324 \times 6}{625}$

$y - 6 = \frac{1944}{625}$

$y - 6 = 3.1104$

$y = 6 + 3.1104$

$y = 9.1104\text{ m}$

Step 4: Verification of Results

Final Solution: The length of the supporting wire attached to the roadway 18 m from the middle is 9.1104 m.

Step 1: Standardizing the Hyperbola Equation

Given the equation: $5y^2 - 9x^2 = 36$

To convert this into the standard form of a hyperbola, we divide the entire equation by $36$ [Algebraic Property of Equality]:

$\frac{5y^2}{36} - \frac{9x^2}{36} = \frac{36}{36}$

$\frac{y^2}{36/5} - \frac{x^2}{4} = 1$

This is of the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, which represents a hyperbola with its transverse axis along the $y$-axis [Standard Equation of a Hyperbola].

Step 2: Identifying Parameters $a$ and $b$

Comparing the given equation to the standard form:

• $a^2 = \frac{36}{5} \implies a = \frac{6}{\sqrt{5}}$
• $b^2 = 4 \implies b = 2$

Step 3: Calculating Eccentricity ($e$)

The relationship between $a, b,$ and $e$ for a hyperbola is defined as $c^2 = a^2 + b^2$ and $e = \frac{c}{a}$ [Geometric Definition of Hyperbola].

$c^2 = \frac{36}{5} + 4 = \frac{36 + 20}{5} = \frac{56}{5}$

$c = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}}$

$e = \frac{c}{a} = \frac{2\sqrt{14}/\sqrt{5}}{6/\sqrt{5}} = \frac{2\sqrt{14}}{6} = \frac{\sqrt{14}}{3}$

Step 4: Determining Coordinates and Latus Rectum

• Vertices: Since the transverse axis is vertical, vertices are at $(0, \pm a)$.
  Vertices: $(0, \pm \frac{6}{\sqrt{5}})$ or $(0, \pm \frac{6\sqrt{5}}{5})$.
• Foci: Foci are located at $(0, \pm c)$.
  Foci: $(0, \pm \frac{2\sqrt{14}}{\sqrt{5}})$ or $(0, \pm \frac{2\sqrt{70}}{5})$.
• Length of Latus Rectum: The formula is $\frac{2b^2}{a}$.
  Length $= \frac{2(4)}{6/\sqrt{5}} = \frac{8\sqrt{5}}{6} = \frac{4\sqrt{5}}{3}$.

Step 5: Visual Representation

Final Solution: Vertices: $(0, \pm \frac{6\sqrt{5}}{5})$; Foci: $(0, \pm \frac{2\sqrt{70}}{5})$; Eccentricity: $\frac{\sqrt{14}}{3}$; Length of Latus Rectum: $\frac{4\sqrt{5}}{3}$

Step 1: Analyzing the Geometric Parameters

Given the ends of the major axis at $(0, \pm\sqrt{5})$ and the ends of the minor axis at $(\pm 1, 0)$, we observe the following:

• Since the major axis lies on the $y$-axis (the $x$-coordinates of the major axis endpoints are zero), the ellipse is a vertical ellipse [Per the standard form of an ellipse centered at the origin].
• The standard equation for a vertical ellipse centered at the origin is: $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a > b$.

Step 2: Identifying the Semi-Axes Lengths

The distance from the center $(0,0)$ to the vertex $(0, \pm a)$ provides the semi-major axis $a$.

• The major axis endpoints are $(0, \pm\sqrt{5})$, thus $a = \sqrt{5}$. Therefore, $a^2 = (\sqrt{5})^2 = 5$.
• The minor axis endpoints are $(\pm 1, 0)$, thus the semi-minor axis $b = 1$. Therefore, $b^2 = (1)^2 = 1$.

Step 3: Constructing the Ellipse Equation

Substituting the values of $a^2$ and $b^2$ into the standard equation:

$\frac{x^2}{1} + \frac{y^2}{5} = 1$

Step 4: Geometric Visualization

Analytical Summary

Final Solution: The equation of the ellipse is $x^2 + \frac{y^2}{5} = 1$ or $5x^2 + y^2 = 5$.

Step 1: Theoretical Foundation of Point-Circle Relationship

To determine the position of a point $P(x_1, y_1)$ relative to a circle defined by the equation $x^2 + y^2 = r^2$, we calculate the power of the point with respect to the circle. Given the standard equation of a circle centered at the origin $(0,0)$ with radius $r$, the condition is defined by the distance $d$ from the origin to the point $P$.

• If $x_1^2 + y_1^2 < r^2$: The point lies inside the circle.
• If $x_1^2 + y_1^2 = r^2$: The point lies on the circle.
• If $x_1^2 + y_1^2 > r^2$: The point lies outside the circle.

Step 2: Substitution of Given Variables

Given the circle equation: $x^2 + y^2 = 25$, where $r^2 = 25$.

Given the point $P(-2.5, 3.5)$, where $x_1 = -2.5$ and $y_1 = 3.5$.

Substitute these values into the expression $S = x_1^2 + y_1^2$:

$S = (-2.5)^2 + (3.5)^2$

Step 3: Algebraic Calculation

Calculating the squares of the coordinates [Per arithmetic square property $(a)^2 = a \cdot a$]:

$(-2.5)^2 = 6.25$

$(3.5)^2 = 12.25$

Summing the squares: $6.25 + 12.25 = 18.5$

Step 4: Comparison with $r^2$

Comparing the calculated value $18.5$ to the constant $r^2 = 25$:

$18.5 < 25$

Since the sum of the squares of the coordinates is strictly less than the square of the radius, the distance from the origin to the point is less than the radius of the circle [Per distance formula $d = \sqrt{x^2 + y^2} < r$].

Visual Representation

Final Solution: Since $(-2.5)^2 + (3.5)^2 = 18.5$, which is less than 25, the point (-2.5, 3.5) lies inside the circle $x^2 + y^2 = 25$.

Step 1: Identification of Hyperbola Orientation and Given Parameters

Given the foci at $(0, \pm 13)$, we observe that the foci lie on the $y$-axis. [Per the standard form of a hyperbola with a vertical transverse axis, the equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$].

• Foci coordinates: $(0, \pm c) = (0, \pm 13) \implies c = 13$.
• Length of the conjugate axis: $2b = 24 \implies b = 12$.

Step 2: Calculating the Semi-Transverse Axis ($a$)

For any hyperbola, the relationship between the semi-axes and the distance to the foci is governed by the equation $c^2 = a^2 + b^2$. [Per the fundamental hyperbola identity].

Substituting the known values:

$13^2 = a^2 + 12^2$

$169 = a^2 + 144$

$a^2 = 169 - 144$

$a^2 = 25$

$a = 5$

Step 3: Constructing the Equation

With $a^2 = 25$ and $b^2 = (12)^2 = 144$, and identifying the vertical orientation (transverse axis on the $y$-axis), we substitute these into the standard form:

$\frac{y^2}{25} - \frac{x^2}{144} = 1$

Step 4: Visual Representation

Step 5: Tabular Summary of Properties

Final Solution: The equation of the hyperbola is $\frac{y^2}{25} - \frac{x^2}{144} = 1$.