Tangent Line Calculation for Ellipse (x^2 + 7y^2 = 8) at Point (3,0)

[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.

 

[sutupidmath]

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!

 

[sutupidmath]

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!

 

[D H]

Slope-intercept form: y=mx+b

What kind of line can't be described in this form?

 

[Rossinole]

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.

 

[D H]

Try finding dx/dy instead.

 

[sutupidmath]

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

 

[Rossinole]

Try finding dx/dy instead.

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?

 

[D H]

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.

 

[Rossinole]

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

 

[sutupidmath]

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!

 

[D H]

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

 

[Rossinole]

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?

 

[D H]

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

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

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

 

[Rossinole]

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?

 

[D H]

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.

 

[Adem Meta]

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.

 

[Adem Meta]

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.