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In summary, the conversation discusses an equation and its representation as a circle or an ellipse. The equation, x/2h+y/2k=1, is mistakenly thought to represent an ellipse, but it actually represents a straight line. To represent an ellipse, the equation should be written as x^2/h + y^2/k = 1. The conversation also touches on the relationship between the center of an ellipse and the slope of two lines that form a right angle.

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- #2

Mentor

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There are no squares in the equation. It is a straight line.Suyash Singh said:this represents an elipse

Did you draw a sketch?

- #3

Mentor

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Here's what you wrote:Suyash Singh said:although i did this for the second equation,

x/2h+y/2k=1

$$\frac x 2 h + \frac y 2 k = 1$$

Here's what you meant:

$$\frac x {2h} + \frac y {2k} = 1$$

When you wrote those fractions on one line, the rules of PEMDAS dictate that the fractions x/2 and y/2 are multiplied by h and k, respectively. If you write them on one line, you need parentheses, like this x/(2h) + y/(2k) = 1.

As already noted, this equation is not the equation of an ellipse.

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Homework Helper

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Not all that easy question, especially as you have to read it half a dozen times to know what it actually is, but it's usually a help to know the answer, which apparently you do.

With that advantage you could think, hmm, Lines coming from points of intersection of a line with a circle that make a right angle – remember some well-known special case?

Can think if you make (x, y) = (0, 0) what happens to your equation to for the circle? That fits your case 2 and the circle going through the origin.

In which case right angle comes from lines from one point on the circumference to other two points on the cIrcumference... again remind of anything?

Centre is the point (k, h).

Line from origin to centre would have slope h/k.

Line at right angles to that would have slope -k/h.

Would have equation kx + hy = something.

Ask if it goes through the centre - it all falls out.

Makes sense, you just have to be able to present it deductively starting from the question not the answer. Hope that helps

With that advantage you could think, hmm, Lines coming from points of intersection of a line with a circle that make a right angle – remember some well-known special case?

Can think if you make (x, y) = (0, 0) what happens to your equation to for the circle? That fits your case 2 and the circle going through the origin.

In which case right angle comes from lines from one point on the circumference to other two points on the cIrcumference... again remind of anything?

Centre is the point (k, h).

Line from origin to centre would have slope h/k.

Line at right angles to that would have slope -k/h.

Would have equation kx + hy = something.

Ask if it goes through the centre - it all falls out.

Makes sense, you just have to be able to present it deductively starting from the question not the answer. Hope that helps

Last edited:

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Suyash Singh said:View attachment 225914

I have no idea what to do please help me.

although i did this for the second equation,

x/2h+y/2k=1

this represents an elipse

first equation is circle

You wrote

$$\frac{x}{2} h + \frac{y}{2} k = 1,$$

and then claimed it represents an ellipse. It does not: it represents a straight line. If you really want an ellipse you need to write

$$\frac{x^2}{h} + \frac{y^2}{k} = 1$$.

The equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

To find the center and radius of a circle, you can rearrange the equation (x - h)^2 + (y - k)^2 = r^2 to solve for (h,k) and r. The values of (h,k) will give you the coordinates of the center, and r will give you the radius.

A circle is a special case of an ellipse, where the length of both the major and minor axes are equal. In an ellipse, the length of the major axis is longer than the length of the minor axis.

The foci of an ellipse can be found using the equation c^2 = a^2 - b^2, where c is the distance from the center to the foci, a is the length of the semi-major axis, and b is the length of the semi-minor axis.

No, the equation for a circle cannot be used to graph an ellipse. The equation for an ellipse is (x - h)^2/a^2 + (y - k)^2/b^2 = 1, where (h,k) is the center of the ellipse, a is the length of the semi-major axis, and b is the length of the semi-minor axis.

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