Archive for the ‘Conic Sections’ Category

Problem 1085

Determine the constant that should be added to the binomial so that it becomes a perfect square trinomial. Then, write and factor the trinomial.

x2 – 14x

 

Solution : –

 

If  x2 + bx is a binomial, then by adding (\frac{b}{a})^{2} , which is the square of half the coefficient of x, a perfect square trinomial will result.

x + bx +(\frac{b}{a})^{2} = (x+\frac{b}{a})^{2}

To complete the square, add the square of half the coefficient of x.

The coefficient of x is -14, so half the coefficient is –\frac{14}{2} = -7.

The square of half the coefficient of x, -7x is (-7)2 = 49.

To write the trinomial, add the constant 49 to the original expression. The resulting  perfect .

To write the trinomial, add the constant 49 to the original expression. The resulting perfect square trinomial is x2 – 14x + 49.

Finally, factor the trinomial x2 – 14x + 49.

x2 -14x + 49  =  (x – 7)2

 

Problem 1084

Find the vertex, focus, and directrix of the following parabola. Then draw the graph.

(x – 2)2 = -3 (y + 1)

g2

Solution:-

 

The two standard forms of the equation for a parabola are (x – h)2 = 4p(y – k) or

(y – k)2 = 4p(x – k).

Since the parabola is already in one of the standard forms, you can determine the values of h, k, and p, and form these values determine the vertex, focus, and directrix.

h = 2, k = -1, and p = –\frac{3}{4}

Remember, for a parabola written in the form (x – h)2 = 4p(y – k), the vertex is (h , k).

Thus , the vertex of this parabola is (2,-1).

The axis of symmetry is vertical. The focus is –\frac{3}{4} units away from the vertex, in the direction of the axis of symmetry.

Thus, it will be \frac{3}{4} units below the vertex.

The focus is (2,-\frac{7}{4}).

The directrix will be \frac{3}{4} units above the vertex.

The equation for the directrix is

y = –\frac{1}{4}

With this information, graph the parabola. First, graph the vertex, focus, and axis of symmetry.

Next, fill in the graph of the parabola.

 

 

Problem 1083

Find the equation if the parabola determined by the given information.

Focus (3,5), directrix x = -1.

 

Solution:-

 

Notice that the directrix is a vertical line. Since the axis of symmetry is perpendicular to the directrix, then it is a horizontal line.

The parabola with a horizontal axis of symmetry will have directrix x = h – p, focus (h +p, k), and standard equation (y – k)2 = 4p(x  – h).

Since the directrix of this parabola is x = -1, then – 1 = h – p.

Since the x – value of the focus is 3, then 3 = h + p.

Find h and p using any method for solving a system of equations. Using the elimination method, eliminate p and solve for h.

h = 1

Substitute the value for h into one of the equation to solve for p.

p = 2

k is the y-value of the focus. So, k = 5.

Substitute the values for h, p, and k into the standard equation of a parabola.

(y – 5)2 = 4 *2(x – 1)

(y – 5)2 = 8(x – 1)

 

 

Problem 717

Find the vertex of the parabola x = 3y2 + 6y+1

 

Solution:-

 

y coordinate =\frac{ -b}{2a}

= \frac{-6}{(2*3)}

= -1
x coordinate = 3(-1)^{2}+6(-1)+1

=3-6+1

=-2
So vertex of the parabola is (-2,-1)

 

Problem 708

Which of the following is another representation of the polar coordinate

P = (8,-45°)

 

Solution:-

 

If point  P =  (r,\theta) , it can also be written as (r,\theta+2\pi k) or (-r,\theta+(2k+1)\pi ) where k is any integer.

(-8,135°)

 

Problem 704

Write the following complex number in polar form. Express all angles in degrees rounded to the nearest tenth.

-3

 

Solution:-

 

-3+0i

 r = \sqrt{x^{2}+y^{2}}

r =\sqrt{-3^{2}}

r = 3

θ = tan^{-1}(\frac{y}{x})

θ = tan^{-1}(\frac{0}{-3})

θ = 0°

polar form (3,0°)

 

Problem 702

Convert the rectangular coordinates (4,4i\sqrt{3})into polar coordinates.

 

Solution:-

 

r^{2} = x^{2} + y^{2}

r^{2 }= 4^{2} + (4sqrt(3))^{2}

{r^2} = 16+48

r^{2} = 64

r = 8

Now  we are find θ =tan^{-1}\frac{y}{x}

= tan^{-1}\frac{4\sqrt{3}}{4}

θ = 60°

So the poler coordinate (8, 60°)

 

Problem 694

You want to make an investment in a continuously compounding account over a period of ten years. What interest rate is required for your investment to double in that time period? Round the logarithm value to the nearest hundredth and the answer to the nearest tenth.

 

Solution:-

 

t =10 years

let initial investment = x

and after 10 years investment will be 2x

2x = x*e^(r/100*10)

r= 6.9 %

 

Problem 525

 

What is the equation of the ellipse with vertices at (-25, 0), (25, 0) and co-vertices (0, -15), (0, 15)?

 

Solution:-

 

Use the standard equation   \frac{ x ^2}{b^2}+\frac{y^2}{a^2} = 1

 

\frac{ x ^2}{625}+\frac{y^2}{225} = 1

Problem 524

 

What is the equation of the ellipse with foci (0, 4), (0, -4) and vertices (0, 8), (0, -8)?

 

Solution:-

 

Use the standard equation   \frac{ x ^2}{b^2}+\frac{y^2}{a^2} = 1

 

\frac{ x ^2}{48}+\frac{y^2}{64} = 1