CE Board Exam Randomizer

Question 1

Parabola

A parabola is the locus of all points equidistant from a fixed point (focus) and a fixed line (directrix).

Standard forms:

Horizontal axis:

$$ (y-k)^2 = 4a(x-h) $$

Vertical axis:

$$ (x-h)^2 = 4a(y-k) $$

(h,k) is the vertex of the parabola

Answer

Distance From the Center of a Parabolic Dish to Its Focus – Diagram

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:

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

Answer

Focus and Directrix of the Parabola – Diagram

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} $

Answer

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

Solve for $y$:

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

Final Answer:

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

Answer

Equation of a Suspension Bridge Cable and Its Height 30 m From the Center – Diagram

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: $17.5\ \text{m}$

Answer

Ceiling Height in a Parabolic Hut – Diagram

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:

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

Answer

Equation of a Parabola in the Form $y = ax^2 + bx + c$ – Diagram

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:

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

Answer

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

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

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

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

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

$\boxed{k = 7}$

Answer

Vertex of the Parabola $y = -x^2 - 2x + 8$ – Diagram

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:

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

Answer

Equation of a Parabola With Vertex $V(2,3)$ and Passing Through $(5,1)$ – Diagram

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:

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

Answer

Find the Vertex of the Parabola $y = 2x^2 - 6x + 4$ – Diagram

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:

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

Answer

Find the Equation, Focus, and Latus Rectum of a Parabola – Diagram

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:

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

Answer

Convert the Cartesian equation to polar form using $x=r\cos\theta$ and $y=r\sin\theta$.
$x^2=8(2-y)$
$r^2\cos^2\theta=16-8r\sin\theta$
Using $\cos^2\theta=1-\sin^2\theta$, this is satisfied by $r=\frac{4}{1+\sin\theta}$:
$r^2\cos^2\theta=\frac{16(1-\sin^2\theta)}{(1+\sin\theta)^2}=\frac{16(1-\sin\theta)}{1+\sin\theta}$
$8(2-r\sin\theta)=16-\frac{32\sin\theta}{1+\sin\theta}=\frac{16(1-\sin\theta)}{1+\sin\theta}$
$\boxed{r=\frac{4}{1+\sin\theta}}$

Answer

The given points $(-3,2)$ and $(-3,-6)$ are $8$ units apart on the line $x = -3$.
If they are a side (length 8), the other pair is at $x = 5$ or $x = -11$: giving $(5,2)$, $(5,-6)$, $(-11,2)$, $(-11,-6)$.
If they are a diagonal, the center is $(-3,-2)$ and the other diagonal (horizontal, half-length 4) gives $(1,-2)$ and $(-7,-2)$.
Thus $(-7,-2)$, $(5,2)$, and $(-11,-6)$ are all possible, but $(1,2)$ fits none.
$\boxed{(1,\ 2) \text{ cannot be a vertex}}$

Answer

Complete the square in $x$:
$x^2 + 4x = 6y + 14$
$(x + 2)^2 - 4 = 6y + 14$
$(x + 2)^2 = 6(y + 3)$
Vertex at $(h, k)$:
$\boxed{(-2,\ -3)}$

Answer

In vertex form $y = (x - h)^2 + k$, the vertex is $(h, k)$.
From $y = (x + 1)^2 + 7$: $h = -1$, $k = 7$.
$\boxed{(-1,\ 7)}$

Answer

Vertex x-coordinate: $h = -\frac{b}{2a} = -\frac{8}{2(1)} = -4$.
Then $y = (-4)^2 + 8(-4) + 63 = 16 - 32 + 63 = 47$.
$\boxed{(-4,\ 47)}$

Answer

With vertex at the origin and horizontal axis, $y^2 = 4ax$. Through $(3, 4)$:
$16 = 4a(3) \Rightarrow 4a = \frac{16}{3}, \quad a = \frac{4}{3}$
The area between the parabola and its latus rectum (from vertex to $x = a$) is:
$A = \int_0^a 2\sqrt{4ax}\,dx = \frac{8}{3}a^2 = \frac{8}{3}\left(\frac{4}{3}\right)^2$
$\boxed{A = 4.74}$

Answer

The distance from the focus $(4, -3)$ to the directrix $y = 5$ is:
$|{-3} - 5| = 8$
This distance equals $2a$, so $a = 4$.
Length of the latus rectum:
$LR = 4a = 4(4)$
$\boxed{LR = 16}$

Question 2

Distance From the Center of a Parabolic Dish to Its Focus

The interior of a satellite TV antenna is a dish shaped like a finite paraboloid. The dish has a diameter of 12 meters and is 2 meters deep. Find the distance from the center of the dish to the focus.

Question 3

Focus and Directrix of the Parabola

Find the focus and directrix of the parabola whose equation is $y = x^2$

Question 4

Height of the Gateway Arch at a Point 100 ft From Its Foundation

The shape of the Gateway Arch in St. Louis is approximately a parabola with a base of 630 ft and a height of 630 ft. How high is the arch 100 feet from its foundation?

Question 5

Equation of a Suspension Bridge Cable and Its Height 30 m From the Center

The cables of a horizontal suspension bridge are supported by two towers 120 m apart and 40 m high. If the cable is 10 m above the floor of the bridge at the center and the midpoint of the bridge is taken as the origin, find the equation of the parabola and the vertical distance above the bridge floor to the cable at a distance of 30 m from the center.

Question 6

Ceiling Height in a Parabolic Hut

A construction worker built a parabolic hut which is 16.1 m wide at the base and 12.4 m high at the center. How high above the base should a ceiling 12.2 m wide be constructed?

Question 7

Equation of a Parabola in the Form $y = ax^2 + bx + c$

A parabola has vertex $V(-4, 2)$ and directrix $y = 5$. Express the equation of the parabola in the form $y = ax^2 + bx + c$.

Question 8

Solving for $k$ in the Parabola $y = a(x + h)^2 + k$

A parabola has an equation $y = 2x^2 + 4x + 5$ and an equivalent equation in the form $y = a(x + h)^2 + k$. Solve for $k$.

Question 9

Vertex of the Parabola $y = -x^2 - 2x + 8$

A parabola has an equation $y = -x^2 - 2x + 8$. Find the coordinates of the vertex of the parabola.

Question 10

Equation of a Parabola With Vertex $V(2,3)$ and Passing Through $(5,1)$

Find an equation of a parabola that has vertex $V(2,3)$, has a vertical axis (note: this means a vertical axis of symmetry), and passes through the point $(5,1)$.

Question 11

Find the Vertex of the Parabola $y = 2x^2 - 6x + 4$

Question 12

Find the Equation, Focus, and Latus Rectum of a Parabola

Find an equation of a parabola with vertex $(2, -3)$ and directrix $y = 4$. Determine also the focus of the parabola and its latus rectum.

Question 13

Question Bank: q246

MSTE - Analytic Geometry / Polar Equation / Engr. Janclyde Espinosa (Clidez)

Find a polar equation that has the same graph as the parabola x²= 8(2-y)

Answer:

Question 14