CE Board Exam Randomizer

Since the point $P(x,y)$ is equidistant from $(1,4)$ and $(5,2)$:

$$ \sqrt{(x-1)^2 + (y-4)^2} = \sqrt{(x-5)^2 + (y-2)^2} $$ Square both sides: $$ (x-1)^2 + (y-4)^2 = (x-5)^2 + (y-2)^2 $$ Expand: $$ x^2 - 2x + 1 + y^2 - 8y + 16 = x^2 - 10x + 25 + y^2 - 4y + 4 $$ Combine like terms: $$ 10x - 2x - 8y + 4y - 12 = 0 $$ Simplify: $$ 8x - 4y - 12 = 0 $$ Divide by 4: $$ 2x - y - 3 = 0 $$

The locus is the line:

$$ \boxed{2x - y - 3 = 0} $$

The condition is:

$$ d_1 = \tfrac12 d_2 $$ Distance to the point $(0,1)$: $$ d_1 = \sqrt{(x-0)^2 + (y-1)^2} $$ Distance to the line $y=4$: $$ d_2 = |4 - y| $$ Impose the condition: $$ \sqrt{(x)^2 + (y-1)^2} = \tfrac12 (4 - y) $$ Square both sides: $$ x^2 + (y-1)^2 = \tfrac14 (4 - y)^2 $$ Expand: $$ x^2 + y^2 - 2y + 1 = \tfrac14 (16 - 8y + y^2) $$ Multiply both sides by 4: $$ 4x^2 + 4y^2 - 8y + 4 = 16 - 8y + y^2 $$ Simplifying, we obtain the equation of an ellipse. $$ 4x^2 + 3y^2 = 12 $$ Divide by 12: $$ \frac{x^2}{3} + \frac{y^2}{4} = 1 $$ Thus the ellipse has center $(0,0)$, with: $$ a = 2,\quad b = \sqrt{3} $$ Compute $c$: $$ c^2 = a^2 - b^2 = 4 - 3 = 1 $$ $$ c = 1 $$ Eccentricity: $$ e = \frac{c}{a} = \frac12 $$

Therefore, the eccentricity is:

$$ \boxed{\frac12} $$

Observation: The equation is quadratic in $x$ but only first degree in $y$. A curve of the form $$Ax^2 + Ey + F = 0$$ is a parabola. Hence, the locus is a parabola.

Step 1: Subtract the equations of the two circles

C₁:  x² + y² + 2x + 4y − 3 = 0  
C₂:  x² + y² − 8x − 6y + 7 = 0
--------------------------------
       10x + 10y − 10 = 0
    

Simplifying:

$\boxed{x + y = 1}$

This line is the common tangent at the point of contact.

When two circles are tangent internally or externally, their point of tangency lies on a line that is perpendicular to the line joining the centers. The key idea: The difference of the equations of two circles gives the radical axis—the locus of points having equal power with respect to both circles.

If two circles intersect at two points, the radical axis is their common chord. But if the circles touch at exactly one point (tangent), the radical axis collapses to the common tangent at the point of contact.

Therefore, since subtracting C₂ from C₁ yields a linear equation, it must represent the common tangent.

Step 1: Determine the radius using the tangency to the y-axis

The distance from the center $(2,3)$ to the y-axis $(x = 0)$ is: $$ r = |2| = 2 $$

Step 2: Use the right triangle formed by the center and point of tangency

Let the point of tangency be $P(3, y)$. The horizontal distance from the center to $P$ is: $$ 3 - 2 = 1 $$

Let the vertical distance be $h$. Using the Pythagorean relationship: $$ h^2 + 1^2 = r^2 $$ $$ h^2 + 1 = 4 $$ $$ h^2 = 3 $$ $$ h = \sqrt{3} $$

Step 3: Compute the y-coordinate

$$ y = 3 + h $$ $$ y = 3 + \sqrt{3} $$

Final Answer:

$$ \boxed{y = 3 + \sqrt{3}} $$

Substitute the point (1,0) into the circle equation:

\[ (1 + 2k)^2 + (0 - 3k)^2 = 10 \]

Expand each term:

\[ (1 + 2k)^2 = 1 + 4k + 4k^2 \] \[ (-3k)^2 = 9k^2 \]

Sum them:

\[ 1 + 4k + 4k^2 + 9k^2 = 10 \]

Combine like terms:

\[ 13k^2 + 4k + 1 = 10 \]

Move 10 to the left:

\[ 13k^2 + 4k - 9 = 0 \]

Solve the quadratic:

\[ k = \frac{-4 \pm \sqrt{4^2 - 4(13)(-9)}}{2(13)} \] \[ k = \frac{-4 \pm \sqrt{16 + 468}}{26} \] \[ k = \frac{-4 \pm \sqrt{484}}{26} \] \[ k = \frac{-4 \pm 22}{26} \]

Thus, the two possible values are:

\[ k = \frac{18}{26} = \frac{9}{13}, \qquad k = \frac{-26}{26} = -1 \]

Final Answer:

\[ \boxed{k = -1 \quad \text{or} \quad k = \frac{9}{13}} \]

Step 1: Rewrite the circle in standard form

$$x^2 + y^2 - 16y = 0$$ Complete the square: $$x^2 + (y^2 - 16y + 64) = 64$$ $$(y - 8)^2 + x^2 = 8^2$$ The center is $(0,8)$ and the radius is $$r = 8.$$

Step 2: Find the distance from the point to the center

$$d_1 = \sqrt{(8 - 0)^2 + (10 - 8)^2}$$ $$d_1 = \sqrt{64 + 4} = \sqrt{68} = 8.25$$

Step 3: Farthest distance = center-to-point distance + radius

$$d_2 = d_1 + r$$ $$d_2 = 8.25 + 8 = 16.25$$

Final Answer:

$$\boxed{16.25}$$

A vertical parabola opening to the right has the form $$y^2 = 4ax$$

The radius of the dish is half the diameter: $$6 \text{ meters}$$ The depth of the dish is $$2 \text{ meters}$$

Substitute the point on the rim of the parabola: $$y = 6, \quad x = 2.$$

Plug into the parabola equation: $$(6)^2 = 4a(2)$$ $$36 = 8a$$ $$a = 4.5.$$

Final Answer:

$$\boxed{4.5 \text{ meters}}$$

Write the parabola in the form $y = 4ax^2$

Compare coefficients: $4a = 1$
$a = \frac{1}{4}$

The focus of $y = 4ax^2$ is $(0, a) = \left(0, \frac{1}{4}\right)$

The directrix is $y = -a = -\frac{1}{4}$

Final Answers:

$ \text{Focus: } \left(0, \frac{1}{4}\right) $ $ \text{Directrix: } y = -\frac{1}{4} $

Using similar triangles: $\dfrac{315^2}{630} = \dfrac{215^2}{630 - y}$

Solve for $y$:

$630 - y = 293.49$ $y = 336.50$

Final Answer:

$\boxed{337\ \text{ft}}$

Vertex at the origin gives the form $x^2 = 4a(y - 10)$

At the tower location, $x = 60$ and $y = 40$:

$60^2 = 4a(40 - 10)$ $3600 = 120a$ $a = 30$

Substitute $a$ back into the parabola:

$x^2 = 120(y - 10)$ $x^2 = 120y - 1200$

To find the height at $x = 30$:

$30^2 = 120y - 1200$ $900 = 120y - 1200$ $120y = 2100$ $y = 17.5\ \text{m}$

Final Answers:

Equation of the cable: $x^2 = 120(y - 10)$
Height at 30 m from center: $17.5\ \text{m}$

By the squared property of a parabola: $\dfrac{8.05^2}{12.4} = \dfrac{6.1^2}{y}$

Solve for $y$:

$y = 7.12$

Ceiling height above the base: $h = 12.4 - 7.12$ $h = 5.28\ \text{m}$

Final Answer:

$\boxed{5.28\ \text{m}}$

A vertical parabola opening downward has the form $$ (x - h)^2 = -4a(y - k) $$
This opens downward because the parabola always faces away from the directrix. Since the directrix is above y=2 (vertex), the parabola is concave downward.

Substituting the vertex $(-4, 2)$: $$ (x + 4)^2 = -4a(y - 2) $$

Distance from vertex to directrix: $$ 5 - 2 = 3 $$ so $a = 3$.

Substitute $a = 3$: $$ (x + 4)^2 = -4(3)(y - 2) $$

Expand:

$$ x^2 + 8x + 16 = -12y + 24 $$

Rearrange:

$$ 12y = -x^2 - 8x + 8 $$

Divide by 12:

$$ y = -\frac{x^2}{12} - \frac{2}{3}x + \frac{2}{3} $$

Final Answer:

$$\boxed {y = -\frac{x^2}{12} - \frac{2}{3}x + \frac{2}{3}} $$

$y = 2x^2 + 4x + 5$

$y = 2(x^2 + 2x) + 5$

$y = 2(x^2 + 2x + 1) + 5 - 2$

$y = 2(x + 1)^2 + 7$

Compare to $y = a(x + h)^2 + k$:

$\boxed{k = 7}$

For a parabola $y = ax^2 + bx + c$, the $x$-coordinate of the vertex is $$x = -\frac{b}{2a}$$

Here, $a = -1$ and $b = -2$.

$$x = -\frac{-2}{2(-1)}$$ $$x = -1$$

Substitute $x = -1$ into the equation:

$y = -(-1)^2 - 2(-1) + 8$ $y = -1 + 2 + 8$ $y = 9$

Final Answer:

$$\boxed{(-1,\, 9)}$$

Vertex form for a vertically oriented parabola: $$ (x - h)^2 = -4a(y - k) $$

Here, $h = 2$ and $k = 3$.

Substitute the point $(5,1)$: $$ (5 - 2)^2 = -4a(1 - 3) $$

$$ 9 = 8a $$ $$ a = \frac{9}{8} $$

Substitute $a$ back into the vertex form:

$$ y - 3 = -\frac{2}{9}(x - 2)^2 $$

Final equation:

$$ y = -\frac{2}{9}(x - 2)^2 + 3 $$

Final Answer:

$$ \boxed{\,y = -\frac{2}{9}(x - 2)^2 + 3\,} $$

For a parabola $y = ax^2 + bx + c$: $$x_\text{vertex} = -\frac{b}{2a}$$

Here, $a = 2$ and $b = -6$.

Since a>0 (the parabola opens upward)

$$x = -\frac{-6}{2(2)} = \frac{3}{2}$$

Substitute $x = \tfrac{3}{2}$ into $y = 2x^2 - 6x + 4$:

$$y = 2\left(\frac{3}{2}\right)^2 - 6\left(\frac{3}{2}\right) + 4$$

$$y = 2\left(\frac{9}{4}\right) - 9 + 4$$ $$y = \frac{9}{2} - 5$$ $$y = -\frac{1}{2}$$

Final Answer:

$$\boxed{\left(\frac{3}{2},\ -\frac{1}{2}\right)}$$

The vertex is $(h, k) = (2, -3)$ and the directrix is $y = 4$. The distance from the vertex to the directrix is: $$4 - (-3) = 7$$ So $a = 7$.

Equation of a vertical parabola opening downward: $$ (x - h)^2 = -4a(y - k) $$

Substituting $h = 2$, $k = -3$, and $a = 7$:

$$ (x - 2)^2 = -4(7)(y + 3) $$ $$ (x - 2)^2 = -28(y + 3) $$

Focus:

The focus lies $a$ units below the vertex (downward opening):

$$F(2,\ -3 - 7) = (2,\ -10)$$

Latus Rectum:

$$L = 4a = 4(7) = 28$$

Final Answers:

$$\boxed{(x - 2)^2 = -28(y + 3)}$$ $$\boxed{F(2,-10)}$$ $$\boxed{L = 28}$$

Standard form of a horizontal ellipse centered at the origin: $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

The vertices are at $(\pm 4, 0)$, so $$ a = 4 $$

The foci are at $(\pm 2, 0)$, so $$ c = 2 $$

Relationship for ellipses: $$ a^2 = b^2 + c^2 $$

Substitute values: $$ 4^2 = b^2 + 2^2 $$ $$ 16 = b^2 + 4 $$ $$ b^2 = 12 $$

Final Equation:

$$ \boxed{\frac{x^2}{16} + \frac{y^2}{12} = 1} $$

The ellipse is horizontal and centered at the origin, so its equation is $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

Since the ellipse has an $x$-intercept of $5$, $$ a = 5 $$

Given a focus at $(2,0)$, $$ c = 2 $$

For ellipses: $$ b^2 = a^2 - c^2 $$

Substitute values: $$ b^2 = 25 - 4 $$ $$ b^2 = 21 $$

Final Equation:

$$ \boxed{\frac{x^2}{25} + \frac{y^2}{21} = 1} $$

Start with the ellipse equation: $$9x^2 + 4y^2 = 25$$

Solve for $y$ to obtain the upper and lower halves:

$$4y^2 = 25 - 9x^2$$ $$y^2 = \frac{25 - 9x^2}{4}$$ $$y = \pm \frac{1}{2}\sqrt{25 - 9x^2}$$

Upper half:

$$y = \frac{1}{2}\sqrt{25 - 9x^2}$$

Lower half:

$$y = -\frac{1}{2}\sqrt{25 - 9x^2}$$

Left and Right Halves

Solve for $x$:

$$9x^2 = 25 - 4y^2$$ $$x^2 = \frac{25 - 4y^2}{9}$$ $$x = \pm \frac{1}{3}\sqrt{25 - 4y^2}$$

Left half:

$$x = -\frac{1}{3}\sqrt{25 - 4y^2}$$

Right half:

$$x = \frac{1}{3}\sqrt{25 - 4y^2}$$

Final Answers:

Upper half: $$\boxed{y = \frac{1}{2}\sqrt{25 - 9x^2}}$$ Lower half: $$\boxed{y = -\frac{1}{2}\sqrt{25 - 9x^2}}$$ Left half: $$\boxed{x = -\frac{1}{3}\sqrt{25 - 4y^2}}$$ Right half: $$\boxed{x = \frac{1}{3}\sqrt{25 - 4y^2}}$$

Total major axis length:

$$2a = 410 + 400 + 400 + 2450$$ $$2a = 3660$$ $$a = 1830$$

Distance between center and perigee:

$$c = a - 810$$ $$c = 1830 - 810$$ $$c = 1020$$

Using $c = ae$:

$$1020 = 1830e$$ $$e = 0.557$$

Final Answer:

$$\boxed{e = 0.557}$$

Step 1: Substitute $(4,0)$ into the equation.

$$A(4)^2 + B(0)^2 + F = 0$$ $$16A + F = 0$$ $$\frac{F}{A} = -16$$

Step 2: Substitute $(0,3)$.

$$A(0)^2 + B(3)^2 + F = 0$$ $$9B + F = 0$$ $$B = -\frac{F}{9}$$

Step 3: Divide original equation by $A$.

$$x^2 + \frac{B}{A}y^2 + \frac{F}{A} = 0$$

Substitute $\frac{F}{A} = -16$ and $\frac{B}{A} = \frac{16}{9}$:

$$x^2 + \frac{16}{9}y^2 - 16 = 0$$

Final Answer:

$$\boxed{9x^2 + 16y^2 - 144 = 0}$$

The ellipse equation is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Here, the semi-major axis is $a = 25$ ft and the semi-minor axis is $b = 15$ ft: $$\frac{x^2}{25^2} + \frac{y^2}{15^2} = 1$$

The truck's right side is 14 ft from the centerline, so evaluate the arch height at $x = 14$: $$\frac{14^2}{25^2} + \frac{y^2}{15^2} = 1$$

Solve for $y$: $$y^2 = 154.44$$ $$y = 12.43 \text{ ft}$$

The truck is 12 ft tall, so clearance is: $$h = 12.43 - 12 = 0.43 \text{ ft}$$

Final Answer:

$$\boxed{0.43\text{ ft of clearance}}$$

Complete the square:

$$x^2 - 2x + 1 - 4y^2 - 63 = 1$$ $$(x - 2)^2 - 4y^2 = 64$$ $$\frac{(x - 2)^2}{64} - \frac{y^2}{16} = 1$$

Thus, $$a^2 = 64,\qquad b^2 = 16$$ $$a = 8,\qquad b = 4$$

For a horizontal hyperbola, the asymptote angle satisfies: $$\tan\theta = \frac{b}{a}$$

$$\tan\theta = \frac{4}{8} = \frac{1}{2}$$ $$\theta = 26.6^\circ$$

Final Answer:

$$\boxed{26.6^\circ}$$

Start with the standard hyperbola: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

Substitute the point $\left(\frac{5}{2},\, 3\right)$:

$$ \frac{\left(\frac{5}{2}\right)^2}{a^2} - \frac{3^2}{b^2} = 1 $$ $$ \frac{25}{4a^2} - \frac{9}{b^2} = 1 $$

Since the asymptote has slope $m = 2$, $$m = \frac{b}{a} \quad\Rightarrow\quad b = 2a$$

Substitute $b = 2a$:

$$ \frac{25}{4a^2} - \frac{9}{(2a)^2} = 1 $$ $$ \frac{25}{4a^2} - \frac{9}{4a^2} = 1 $$ $$ \frac{16}{4a^2} = 1 $$ $$ a^2 = 4,\quad a = 2 $$

Then: $$b = 2a = 4$$

The hyperbola becomes: $$ \frac{x^2}{4} - \frac{y^2}{16} = 1 $$

Latus Rectum Length

For a horizontal hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the latus rectum length is: $$L = \frac{2b^2}{a}$$

$$ L = \frac{2(4^2)}{2} = \frac{2 \cdot 16}{2} = 16 $$

Final Answer:

$$\boxed{16}$$

Start with the general equation: $$Ax^2 + By^2 + F = 0$$ Divide by $A$: $$x^2 + \frac{B}{A}y^2 + \frac{F}{A} = 0$$

Substitute the point $(0,3)$:

$$ 0 + \frac{B}{A}(3^2) + \frac{F}{A} = 0 $$ $$ 9\frac{B}{A} + \frac{F}{A} = 0 $$ $$ \frac{F}{A} = -9\frac{B}{A} $$

Substitute the point $(4,6)$:

$$ (4)^2 + \frac{B}{A}(6^2) + \frac{F}{A} = 0 $$ $$ 16 + 36\frac{B}{A} + \frac{F}{A} = 0 $$ Substitute $\frac{F}{A} = -9\frac{B}{A}$: $$ 16 + 36\frac{B}{A} - 9\frac{B}{A} = 0 $$ $$ 16 + 27\frac{B}{A} = 0 $$ $$ \frac{B}{A} = -\frac{16}{27} $$

So the equation becomes: $$ x^2 - \frac{16}{27}y^2 - 9 = 0 $$

Multiply by 27: $$ 27x^2 - 16y^2 - 243 = 0 $$

Conclusion:

The equation has one positive and one negative squared term $$27x^2 - 16y^2 + 144 = 0$$ which is the equation of a hyperbola since A and C have opposite signs in the general form: $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$.

$$\boxed{\text{The curve is a hyperbola.}}$$

For a vertical hyperbola of the form $$\frac{y^2}{a^2} - \frac{(x - h)^2}{b^2} = 1,$$ we identify: $$a^2 = 2,\quad b^2 = 36.$$

So: $$a = \sqrt{2}, \qquad b = 6.$$

The relationship between $a$, $b$, and $c$ for hyperbolas is: $$c^2 = a^2 + b^2.$$

Substitute: $$c^2 = 2 + 36 = 38,$$ $$c = \sqrt{38}.$$

Eccentricity is: $$e = \frac{c}{a} = \frac{\sqrt{38}}{\sqrt{2}}.$$

Simplify: $$e = \sqrt{\frac{38}{2}} = \sqrt{19} \approx 4.4.$$

Final Answer:

$$\boxed{e \approx 4.4}$$

Step 1: Rewrite in standard form.

$$ \frac{x^2}{4} - \frac{y^2}{9} = 1 $$ Thus, $$ a^2 = 4,\quad a = 2 $$ $$ b^2 = 9,\quad b = 3 $$

Step 2: Find the foci.

For hyperbolas: $$ c^2 = a^2 + b^2 = 4 + 9 = 13 $$ $$ c = \sqrt{13} $$ Since the transverse axis is horizontal: $$ F\left(\sqrt{13},\,0\right),\quad F'\left(-\sqrt{13},\,0\right) $$

Step 3: Use the general asymptote formula.

For a hyperbola $$ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, $$ the asymptotes are: $$ y - k = \pm \frac{b}{a}(x - h) $$ Here, the center is $(h,k) = (0,0)$. So: $$ y = \pm \frac{b}{a}x $$ Substitute $a = 2,\ b = 3$: $$ y = \pm \frac{3}{2}x $$

Final Answers:

Foci:

$$ \boxed{F(\sqrt{13},0)},\qquad \boxed{F'(-\sqrt{13},0)} $$

Asymptotes:

$$ \boxed{y = \frac{3}{2}x},\qquad \boxed{y = -\frac{3}{2}x} $$

Since the vertices are $(\pm 3,0)$, the hyperbola is centered at the origin with horizontal transverse axis, so:

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

Given the vertices, $a = 3$ so:

$$ a^2 = 9 $$

Use the point $P(5,2)$:

$$ \frac{5^2}{9} - \frac{2^2}{b^2} = 1 $$ $$ \frac{25}{9} - \frac{4}{b^2} = 1 $$ $$ \frac{25}{9} - 1 = \frac{4}{b^2} $$ $$ \frac{16}{9} = \frac{4}{b^2} $$ $$ b^2 = \frac{9}{4} $$

The standard equation becomes:

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

Multiply through to simplify:

$$ x^2 - 4y^2 = 9 $$

Foci

$$ c^2 = a^2 + b^2 = 9 + \frac{9}{4} = \frac{45}{4} $$ $$ c = \frac{3\sqrt{5}}{2} $$

Thus the foci are:

$$ F\left(\frac{3\sqrt{5}}{2},\,0\right),\qquad F'\left(-\frac{3\sqrt{5}}{2},\,0\right) $$

Asymptotes

General asymptote equation for a hyperbola centered at $(h,k)$:

$$ y - k = \pm \frac{b}{a}(x - h) $$

Here $(h,k) = (0,0)$, $a = 3$, $b = \frac{3}{2}$:

$$ y = \pm \frac{b}{a}x = \pm \frac{\frac{3}{2}}{3}x $$ $$ y = \pm \frac{1}{2}x $$

Final Answers

Equation of hyperbola:

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

Foci:

$$ \boxed{F\left(\frac{3\sqrt{5}}{2},0\right)},\quad \boxed{F'\left(-\frac{3\sqrt{5}}{2},0\right)} $$

Asymptotes:

$$ \boxed{y = \frac{1}{2}x},\qquad \boxed{y = -\frac{1}{2}x} $$

Rewrite the hyperbola in standard form:

$$ 4y^2 - 2x^2 = 1 $$ Divide by 1: $$ \frac{y^2}{\frac{1}{4}} - \frac{x^2}{\frac{1}{2}} = 1 $$ So: $$ a^2 = \frac{1}{4},\qquad b^2 = \frac{1}{2} $$

Compute $a$, $b$, and $c$.

$$ a = \frac{1}{2} $$ $$ b = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$ For hyperbolas: $$ c^2 = a^2 + b^2 $$ $$ c^2 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} $$ $$ c = \frac{\sqrt{3}}{2} $$

Foci: For a vertical hyperbola, the foci are $(0,\pm c)$.

$$ \boxed{(0,\pm \frac{\sqrt{3}}{2})} $$

Asymptotes

General formula for a vertical hyperbola centered at $(h,k)$: $$ y - k = \pm \frac{a}{b}(x - h) $$ Here $(h,k) = (0,0)$, so: $$ y = \pm \frac{a}{b}x $$ Substitute: $$ \frac{a}{b} = \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $$ Thus: $$ \boxed{y = \pm \frac{\sqrt{2}}{2}x} $$

Final Answers

Foci:

$$ \boxed{(0,\pm \frac{\sqrt{3}}{2})} $$

Asymptotes:

$$ \boxed{y = \frac{\sqrt{2}}{2}x},\qquad \boxed{y = -\frac{\sqrt{2}}{2}x} $$

Rewrite the equation in standard hyperbola form:

$$ 4x^2 - y^2 = 16 $$ Divide by 16: $$ \frac{x^2}{4} - \frac{y^2}{16} = 1 $$

This is a hyperbola centered at $(0,0)$ with:

$$ a^2 = 4,\quad a = 2 $$ $$ b^2 = 16,\quad b = 4 $$

For a hyperbola of the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$$ the shortest distance from the center to the curve (closest approach to the nucleus) occurs at the vertex $(\pm a, 0)$.

Thus, the particle comes within

$$ \boxed{a = 2} $$

units of distance from the nucleus.

Differentiation:

$$ 64x^2 + 25y^2 = 1600 $$ Differentiate: $$ 128x + 50y\,y' = 0 $$

Substitute the slope of the parallel chords: $y' = \frac{1}{5}$

$$ 128x + 50y\left(\frac{1}{5}\right) = 0 $$ Simplify: $$ 128x + 10y = 0 $$ Final equation of the diameter: $$ \boxed{64x + 5y = 0} $$

Chords parallel to the line $x - y = 4$ have slope $$ y' = 1 $$

Differentiate the parabola:

$$ y^2 = 8x $$ $$ 2y\,y' = 8 $$

Substitute the slope of the parallel chords:

$$ 2y(1) = 8 $$ Thus, $$ \boxed{y = 4} $$

This horizontal line is the diameter that bisects all chords parallel to $x - y = 4$.