# Tangent lines to an ellipse.

• Rossinole
In summary, the conversation discusses finding the tangent lines to the ellipse x^2 + 7y^2 = 8 at the point (3,0) and the difficulties encountered in finding the slope of the tangent lines. The conversation includes various approaches and equations, such as the slope-intercept form, finding dx/dy, and using the point-slope form. It is eventually determined that there are two tangent lines to the ellipse, and the equation for these lines is xx1/a^2 + yy1/b^2 = 1, where (x1,y1) are the coordinates of the point where the tangent line touches the ellipse. It is also noted that the derivative of the equation of the ellipse does not exist at

#### Rossinole

Find the tangent lines to the ellipse x^2 + 7y^2 = 8 at the point (3,0)

Slope-intercept form: y=mx+b

I know you have to differentiate the equation implicitly to get the slope, but you come across a zero in the denominator and that has me stumped.

Rossinole said:
Find the tangent lines to the ellipse x^2 + 7y^2 = 8 at the point (3,0)

Slope-intercept form: y=mx+b

I know you have to differentiate the equation implicitly to get the slope, but you come across a zero in the denominator and that has me stumped.
Show us what u did? I mean show all of your work, and point out where you're stumpt, so someone will point you on the right direction!

Welcome to pf, by the way! It is a pf's policy not to give answers, but to give hints, so since you are on the homework forum you are supposed to do your homework on your own, while people here will only give hints to you!

Rossinole said:
Slope-intercept form: y=mx+b
What kind of line can't be described in this form?

x^2+7y^2=8 (3,0)

The derivative is 2x + 14y(dy/dx)=0. To get dy/dx by itself, first subtract 2x on both sides and you get 14y(dy/dx) = -2x. Then divide by 14y on both sides and you get (dy/dx)=-2x/14y. Plugging in the points (3,0) into the equation gives you 6/0, which is undefined.

This is where I'm stuck. I know there has to be another way to approach this, I just need someone to point me in the right direction.

for what type of line we say has no slope. Or roughly speaking has a slope of "infinity"?

D H said:

When you solve for dx/dy, you get 2x(dx/dy)+14y=0, 2x(dx/dy)=-14y, (dx/dy)=-14y/2x, (dx/dy)=-14(0)/2(3), (dx/dy)=0/6, (dx/dy)=0. So you end with the tangent line being y=0. There's still supposed to be at least one more line. This, again, is what is confusing me. How would I go about finding more than one tangent line?

There is only one tangent line to an ellipse at any point on the ellipse. An ellipse is a simple curve (it doesn't cross itself; the symbol for infinity or the digit "8" are examples of curve that cross themselves) and it doesn't have a cusp.

I'm pretty sure the point isn't on the ellipse. I'm sorry I guess I should have made that more clear.

Rossinole said:
I'm pretty sure the point isn't on the ellipse. I'm sorry I guess I should have made that more clear.

And you are letting us know just now?
well then the eq of the line is y=mx+b, let this line go throught the point (3,0) so we get

0=3m+b, now the slope of the line at any point in the tangent line as you computed is (dy/dx)=-2x/14y so
0=3(-2x/14y)+b, now this is an eq of the line that goes through (3,0) try to find at what points does this line touch the elipse!

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Doh! We should have seen that. Yes, there are two such lines, neither one is vertical.

sutupidmath said:
And you are letting us know just now?
well then the eq of the line is y=mx+b, let this line go throught the point (3,0) so we get

0=3m+b, now the slope of the line at any point in the tangent line as you computed is (dy/dx)=-2x/14y so
0=3(-2x/14y)+b, now this is an eq of the line that goes through (3,0) try to find at what points does this line touch the elipse!

Shouldn't it be the (dx/dy) in the slope, not (dy/dx)? But my guess for that would be plug in the points into the equation and you get 0=3(6/0)+b, which, you can't do..If you did it (dx/dy), you'd end up with b=0?

It might be easier to work with the point-slope form for a line:

$$\frac {y-y_1}{x-x_1} = m$$

where the reference point $(x_1,y_1)[/tex] is some point on the ellipse. You already have an expression for [itex]m$:

$$\frac{dy}{dx}=-\,\frac{\phantom{1}2x_1}{14y_1}$$

You also know the coordinates of another point on the line: $(3,0)$.See where you can go with this.

Haha, I really don't know where to go with it. Substituting the coordinates of the other point on the line and setting it equal to (dy/dx) is my first guess. 0-y1/3-x1=-(2x/14y). But even then I don't know where to go with that.

Is this in anyway similar to solving for two tangent lines of a parabola?

First, put the expression for the known slope of the line into the equation of the line. Using the point-slope form

$$\frac {y-y_1}{x-x_1} = -\,\frac{x_1}{7 y_1}$$

(I simplified 2/14 to 1/7 in the above). You know an (x,y) pair, (3,0). Put this into the equation. You will be left with a relation between $x_1$ and $y_1$. You already have another relation between these parameters: the equation of the ellipse.

sutupidmath said:
Show us what u did? I mean show all of your work, and point out where you're stumpt, so someone will point you on the right direction!

The equation of a tangent line to ellipse is xx1/a^2+ yy1/b^2=1, where (x1, y1) are the coordinats of the point where the tangent line touch the ellipse in our case (3,0). Then the equation is x=8/3. Another way the equation is y-y1=m(x-x1),we have to find the slope m for that let find the derivative for the equation of the ellipse and you will see that doesn't exist at the point (3,0)because the tangent is a vertical line.

The equation of a tangent line to ellipse is xx1/a^2+ yy1/b^2=1, where (x1, y1) are the coordinats of the point where the tangent line touch the ellipse in our case (3,0). Then the equation is x=8/3. Another way the equation is y-y1=m(x-x1),we have to find the slope m for that let find the derivative for the equation of the ellipse and you will see that doesn't exist at the point (3,0)because the tangent is a vertical line.

## What is a tangent line to an ellipse?

A tangent line to an ellipse is a line that touches the ellipse at exactly one point. This point is called the point of tangency.

## How is a tangent line to an ellipse different from a secant line?

A secant line is a line that intersects an ellipse at two points, while a tangent line only touches the ellipse at one point.

## How do you find the equation of a tangent line to an ellipse?

To find the equation of a tangent line to an ellipse, you need to find the slope of the tangent line at the point of tangency. This can be done by taking the derivative of the ellipse's equation and plugging in the coordinates of the point of tangency. Then, you can use the point-slope form of a line to write the equation of the tangent line.

## Can a tangent line to an ellipse be vertical?

Yes, a tangent line to an ellipse can be vertical. This happens when the ellipse is a circle and the tangent line is perpendicular to the x-axis.

## How many tangent lines can an ellipse have?

An ellipse can have an infinite number of tangent lines. Each point on the ellipse can have its own unique tangent line.