Question about Office Equipment & Supplies

I am not quite sure how the major axis of your hyperbola is directed and i do not know if the lengths you give are measures of the major and minor axes or the measures of the semi-major and semi-minor axes. So I am giving you the equations and the graphs so that you can decide for yourself what is appropriate for your problem.

Major axis parallel to the X-axis

Equation and graph

Center is at x=1 and y=-2, semi-major axis length is a=6, and semi-minor axis length is b=12

Major axis is parallel to the y-axis

Equation

Center is at x=1 and y=-2, semi-major axis length is a=12, and semi-minor axis length is b=6.

I trust you can customize the equations to fit your need.

Posted on Mar 26, 2011

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Posted on Jan 02, 2017

Being parallel to the given line, the equation of the line you are seeking has the same slope, which in this case is **a=1/4.**

So the equation sought is as follows

y=**(1/4)x** +b, where b is to be found.

To find** b**, use the stated fact that the line passes through the point **(x=8, y=-1)**. All that means is that the point **(8,-1)** is on the line whose equation you are looking for. If it is on the line with equation **y=(1/4)x+b**

then its coordinates x=8, and y=-1 satisfy the relation y=(1/4)x+b. In other words, if you substitute 8 for x, and -1 for y, the equality holds true**-1=(1/4)*8 +b**

This gives you a way to find the initial value of the function (the y-intercept b ). Just solve**-1=(1/4)*8 +b** to find b.

I leave this pleasure to you.

So the equation sought is as follows

y=

To find

then its coordinates x=8, and y=-1 satisfy the relation y=(1/4)x+b. In other words, if you substitute 8 for x, and -1 for y, the equality holds true

This gives you a way to find the initial value of the function (the y-intercept b ). Just solve

I leave this pleasure to you.

Jan 15, 2014 | Mathsoft StudyWorks! Mathematics Deluxe...

What is the question? Do you want the distance between the hyperbola and the point? Or did you mean the line y=(1/4)x-2? Do you want the equation of the perpendicular? We can't give you an answer if you don't state a question.

Nov 22, 2013 | Computers & Internet

The x intercept is where your line will cross the x axis. Therefore, the value for y at this point is 0. Substitute 0 for y in this equation and solve for x.

6x + 3(0) = 27

x = 27/6

x = 4 1/2

x intercept = (4 1/2, 0)

Follow the same reasoning for the y intercept.

6(0) + 3y = 27

y = 27/3

y = 9

y intercept = (0,9)

Now plot these two points and draw your line through them.

6x + 3(0) = 27

x = 27/6

x = 4 1/2

x intercept = (4 1/2, 0)

Follow the same reasoning for the y intercept.

6(0) + 3y = 27

y = 27/3

y = 9

y intercept = (0,9)

Now plot these two points and draw your line through them.

May 18, 2012 | SoftMath Algebrator - Algebra Homework...

By definition, the y-intercept is the value of y when x=0. Just set x=0 in the equation and solve for y.

Hence 3(0)+6y=90 or 6y=90. This gives y=90/6=15. The point of intercept of the y-axis is (0,15).

To get the x-intercept, set y= 0 in the equation and solve for x.

3x+6(0)=90 or 3x=90. This gives x=90/3=30.

The point of intercept of the x-axis is (30,0).

Hence 3(0)+6y=90 or 6y=90. This gives y=90/6=15. The point of intercept of the y-axis is (0,15).

To get the x-intercept, set y= 0 in the equation and solve for x.

3x+6(0)=90 or 3x=90. This gives x=90/3=30.

The point of intercept of the x-axis is (30,0).

Jan 20, 2012 | SoftMath Algebrator - Algebra Homework...

The roots are a complex conjugate pair.

Jun 16, 2011 | Office Equipment & Supplies

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Jan 27, 2011 | Texas Instruments TI-84 Plus Calculator

Calculate
the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5,
x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

Oct 20, 2010 | Texas Instruments TI-84 Plus Calculator

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=1, y1=0, x2=0, and x1=-6. You should get a=(1-0)/(0-(-6))=1/6

The y-intercept is the y-cordinate for x=0. Its value is 1.

The equation is then y=(x/6) 1.

The y-intercept is the y-cordinate for x=0. Its value is 1.

The equation is then y=(x/6) 1.

Oct 18, 2010 | Texas Instruments TI-84 Plus Calculator

I'm not sure what your ellipse looks like but the equation for the area of an ellipse is given by:

pi times the long axis times the short axis

pi = 3.1416

The length of the long axis is from the center of the ellipse to the outer edge of the ellipse

The length of the short axis is from the center of the ellipse to the outer edge of the ellipse

Hope this helps Loringh

pi times the long axis times the short axis

pi = 3.1416

The length of the long axis is from the center of the ellipse to the outer edge of the ellipse

The length of the short axis is from the center of the ellipse to the outer edge of the ellipse

Hope this helps Loringh

Nov 01, 2008 | The Learning Company Achieve! Math &...

assuming the question is what is the circle equation?

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

Oct 26, 2008 | Casio FX-115ES Scientific Calculator

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