A parabola is the locus of all points equidistant from a fixed point (focus) and a fixed line (directrix).
Standard forms:
Horizontal axis:
Vertical axis:
(h,k) is the vertex of the parabola
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.
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:
$$\boxed{4.5 \text{ meters}}$$Find the focus and directrix of the parabola whose equation is $y = x^2$
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} $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?
Using similar triangles: $\dfrac{315^2}{630} = \dfrac{215^2}{630 - y}$
Solve for $y$:
$630 - y = 293.49$ $y = 336.50$
Final Answer:
$\boxed{337\ \text{ft}}$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.
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:
$17.5\ \text{m}$
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?
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 parabola has vertex $V(-4, 2)$ and directrix $y = 5$. Express the equation of the parabola in the form $y = ax^2 + bx + c$.
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:
$$\boxed {y = -\frac{x^2}{12} - \frac{2}{3}x + \frac{2}{3}} $$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 $k$.
$y = 2x^2 + 4x + 5$
$y = 2(x^2 + 2x) + 5$
$y = 2(x^2 + 2x + 1) + 5 - 2$
$y = 2(x + 1)^2 + 7$
Compare to $y = a(x + h)^2 + k$:
$\boxed{k = 7}$
A parabola has an equation $y = -x^2 - 2x + 8$. Find the coordinates of the vertex of the parabola.
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:
$$\boxed{(-1,\, 9)}$$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 $(5,1)$.
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:
$$\boxed{\left(\frac{3}{2},\ -\frac{1}{2}\right)}$$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.
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:
$$\boxed{(x - 2)^2 = -28(y + 3)}$$ $$\boxed{F(2,-10)}$$ $$\boxed{L = 28}$$