1. ## Rotated Ellipse

1. Given the equation of the rotated ellipse x^2+xy+y^2=3

a. Find the equation of the normal line (line perpendicular to the tangent line at the given point) to it at x=1

I got 2 points for the problem (1,1) and (1,-2) but you just need one of the normal line. So I did dy/dx=-3y-2x. I picked (1,1) which gives me dy/dx=-5 so the slope of the normal line is 1/5 and the equation of the line would be y-1=(1/5)(x-1). Can someone check my work- I think I'm doing something wrong for some reason.

b. Find the second point on the curve where the normal line is parallel to the one you found in part (a).

I have no idea how to attempt this problem.

c. Find the point(s) on the curve where the tangent line is vertical.

Kinda lost on this problem also...

Thanks!

2. Originally Posted by Linnus
1. Given the equation of the rotated ellipse x^2+xy+y^2=3

a. Find the equation of the normal line (line perpendicular to the tangent line at the given point) to it at x=1

I got 2 points for the problem (1,1) and (1,-2) but you just need one of the normal line. So I did dy/dx=-3y-2x. I picked (1,1) which gives me dy/dx=-5 so the slope of the normal line is 1/5 and the equation of the line would be y-1=(1/5)(x-1). Can someone check my work- I think I'm doing something wrong for some reason.
The equation for you derivative is incorrect. Using implicit differentiation I get:
$\displaystyle x^2+xy+y^2=3$
goes to
$\displaystyle 2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$

Solve for $\displaystyle \frac{dy}{dx}$.

I get that the slopes of the line for x = 1 are $\displaystyle y^{\prime} = -1$ and $\displaystyle y^{\prime} = 0$.

For the -1 slope, the slope of the normal line is +1. I get the line y = x. (See the graph below. This point and line is green.)

For the 0 slope, the slope of the normal is undefined. I get the line x = 1. (This is the blue point and line. Never mind that the blue line also intersects the green point. That's just an artifact of this normal line being parallel to the y-axis.

Originally Posted by Linnus
b. Find the second point on the curve where the normal line is parallel to the one you found in part (a).

I have no idea how to attempt this problem.
Sure you do! It's just that you have to take things step by step.

You know the equation of the normal line, thus you know the slope of the normal line. (Let's just consider the y= x normal line for simplicity.) So you are looking for a point on the ellipse where the normal line has a slope of 1. Thus the tangent line has a slope of -1. Thus:
$\displaystyle \frac{dy}{dx} = -\frac{2x + y}{x + 2y} = -1$

Solve this equation for y and you have the equation of the tangent line that intersects the ellipse where the normal line has a slope of -1. I get the line y = x.

Now you want to find the points where this line intersects the ellipse. So plug y = x into your ellipse formula and solve for x. (There will be typically be two x values also, one of which you already know.) Then you can find the y value for these points by plugging x into y = x. Generally the two points will not be on the same line, as they are in this case.

Originally Posted by Linnus
c. Find the point(s) on the curve where the tangent line is vertical.
This is just the same problem as b) but for the (1, -2) point. Do the same procedure.

-Dan

3. Thank you for helping!
For part b, I got (1,1) for the point, which is the same point as in part a. It asks for a different point @_@ i'm so confused. Shouldn't the other point be (-1,-1) from looking at your graph? Thanks

4. Originally Posted by Linnus
Thank you for helping!
For part b, I got (1,1) for the point, which is the same point as in part a. It asks for a different point @_@ i'm so confused. Shouldn't the other point be (-1,-1) from looking at your graph? Thanks
When you solved $\displaystyle x^2 = 1$ you forgot about the -1 solution.

-Dan

5. oh yea, I forgot that. hmm for part C, can u tell me why do that same thing as part b but for point (1,-2)? Thank you ^^;; sorry for the trouble!

6. Originally Posted by Linnus
oh yea, I forgot that. hmm for part C, can u tell me why do that same thing as part b but for point (1,-2)? Thank you ^^;; sorry for the trouble!
the tangent line is vertical (a vertical line has infinite slope) at the point the derivative is undefined (shoots to infinity) that is, where we divide by zero. find the points x and y for which this happens. there are two for this ellipse

7. so, for part c, i would do x+2y=0 (since that is in the dominator) and i get x= -2, and +2, and plug those values in for the ellipse equation right?

8. Originally Posted by Linnus
so, for part c, i would do x+2y=0 (since that is in the dominator) and i get x= -2, and +2, and plug those values in for the ellipse equation right?
yes, plug those values into the ORIGINAL equation to find the corresponding y-values