MHB Equation of the Normal and Ellipses Questions

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The discussion revolves around solving a problem set involving implicit differentiation and the equations of normals to ellipses. The user correctly derived the derivative dy/dx for part 1a and used it to find the equation for part 1b. For part 2, they noted that the normal line is perpendicular to the tangent, leading to the correct slope calculation. However, there is confusion regarding the second intersection point of the normal line with the ellipse, with the user questioning if the final answer is simply (1, -1). Clarification is needed on how to find this second intersection point.
ardentmed
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Hey guys,

I have a couple of questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:
08b1167bae0c33982682_13.jpg


For 1a, I used implicit differentiation and isolated dy/dx.

This gave me the following answer:

dy/dx = (-3x^2 -4y)/(4x+12y)

Which I then used for 1b and substituted (1,1) for x and y respectively to compute:

y-1= (-7/16) * (x-1)

As for 2, the normal is perpindicular to the tangent line and is thus the negative reciprocal. Thus, if the tangent line's slope is 1, the normal is -1/1= -1.

Thus, I used that for the equation of the line through (-1,1) to get:

(1,-1) (and the other point was already given by the question)

This was obtained by using y=-x as the equation and then substituting x=-y into the x^2 + y^2 = 1 elipse expression to get x^2=1. Thanks in advance.
 
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ardentmed said:
For 1a, I used implicit differentiation and isolated dy/dx.

This gave me the following answer:

dy/dx = (-3x^2 -4y)/(4x+12y)

Which I then used for 1b and substituted (1,1) for x and y respectively to compute:

y-1= (-7/16) * (x-1)

Correct for both a and b part.

ardentmed said:
As for 2, the normal is perpindicular to the tangent line and is thus the negative reciprocal. Thus, if the tangent line's slope is 1, the normal is -1/1= -1.

Thus, I used that for the equation of the line through (-1,1) to get:

(1,-1) (and the other point was already given by the question)

This was obtained by using y=-x as the equation and then substituting x=-y into the x^2 + y^2 = 1 elipse expression to get x^2=1. Thanks in advance.

You've figured out everything correctly so far and now, since solving for $x^2=1$ gives $x=\pm1$, you've to answer the last part of the problem 2 by determining at which point the normal line at $(-1,\,1)$ intersects the ellipse the second time. Can you see how proceed?
 
anemone said:
Correct for both a and b part.
You've figured out everything correctly so far and now, since solving for $x^2=1$ gives $x=\pm1$, you've to answer the last part of the problem 2 by determining at which point the normal line at $(-1,\,1)$ intersects the ellipse the second time. Can you see how proceed?

No, I don't quite get it. Isn't the final answer just (1,-1)?
Thanks in advance.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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