MATH SOLVE

8 months ago

Q:
# The latus rectum of the parabola with the equation 4x 2 - 20x + 24y + 61 = 0 is

Accepted Solution

A:

The latus rectum is 6.

We first find the focal point of the function. To do this, we need the vertex. We will rewrite this in standard form:

4x²-20x+24y+61=0

24y=-4x²+20x-61

y=(-4/24)x²+(20/24)x-(61/24)

y=(-1/6)x²+(5/6)x-(61/24)

We will factor a, -1/6, out of this:

y=(-1/6)(x²-5x+61/4)

The vertex is found by completing the square. To do this, we will find b/2 first:

-5/2

Now we square it:

(-5/2)²=25/4

This is what we add in parentheses to our function:

y = (-1/6)(x²-5x+25/4+61/4)

We must also subtract it:

y = (-1/6)(x²-5x+25/4+61/4-25/4)

The last two numbers will be multiplied by the -1/6 to be removed from parentheses:

y = (-1/6)(x²-5x+25/4)-61/24+25/24

y = (-1/6)(x-5/2)² - 36/24

y = (-1/6)(x-5/2)² - 3/2

Now that it is in vertex form, y=a(x-h)²+k, we see that the vertex, (h, k) is at (5/2, -3/2).

To find the focus, the equation needs to be in the form

(x-h)² = 4p(y-k)²

We know that (h, k) is the vertex:

(x-5/2)² = 4p(y--3/2)²

(x-5/2)² = 4p(y+3/2)²

The difference between this equation and our vertex form is that the k value is moved over with the y, and the value of a in front of x is divided to move it to the y side as well. a = -1/6; dividing by -1/6 is the same as multiplying by -6:

(x-5/2)² = -6(y+3/2)²

This means that 4p = -6, and p = -6/4 = -3/2. The focus is at (h, k+p):

(5/2, -3/2+-3/2) = (5/2, -6/2) = (5/2, -3).

The focal length is the distance from the focus to the vertex. We use the distance formula to find this:

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\=\sqrt{(5/2-5/2)^2+(-3--3/2)^2} \\ \\=\sqrt{0^2+(-3+1.5)^2} \\ \\=\sqrt{0+(-1.5)^2} \\ \\=\sqrt{(-1.5)^2}=1.5 [/tex]

The latus rectum is 4 x the focal length:

4*1.5 = 6

We first find the focal point of the function. To do this, we need the vertex. We will rewrite this in standard form:

4x²-20x+24y+61=0

24y=-4x²+20x-61

y=(-4/24)x²+(20/24)x-(61/24)

y=(-1/6)x²+(5/6)x-(61/24)

We will factor a, -1/6, out of this:

y=(-1/6)(x²-5x+61/4)

The vertex is found by completing the square. To do this, we will find b/2 first:

-5/2

Now we square it:

(-5/2)²=25/4

This is what we add in parentheses to our function:

y = (-1/6)(x²-5x+25/4+61/4)

We must also subtract it:

y = (-1/6)(x²-5x+25/4+61/4-25/4)

The last two numbers will be multiplied by the -1/6 to be removed from parentheses:

y = (-1/6)(x²-5x+25/4)-61/24+25/24

y = (-1/6)(x-5/2)² - 36/24

y = (-1/6)(x-5/2)² - 3/2

Now that it is in vertex form, y=a(x-h)²+k, we see that the vertex, (h, k) is at (5/2, -3/2).

To find the focus, the equation needs to be in the form

(x-h)² = 4p(y-k)²

We know that (h, k) is the vertex:

(x-5/2)² = 4p(y--3/2)²

(x-5/2)² = 4p(y+3/2)²

The difference between this equation and our vertex form is that the k value is moved over with the y, and the value of a in front of x is divided to move it to the y side as well. a = -1/6; dividing by -1/6 is the same as multiplying by -6:

(x-5/2)² = -6(y+3/2)²

This means that 4p = -6, and p = -6/4 = -3/2. The focus is at (h, k+p):

(5/2, -3/2+-3/2) = (5/2, -6/2) = (5/2, -3).

The focal length is the distance from the focus to the vertex. We use the distance formula to find this:

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\=\sqrt{(5/2-5/2)^2+(-3--3/2)^2} \\ \\=\sqrt{0^2+(-3+1.5)^2} \\ \\=\sqrt{0+(-1.5)^2} \\ \\=\sqrt{(-1.5)^2}=1.5 [/tex]

The latus rectum is 4 x the focal length:

4*1.5 = 6