# Proof using implicit differentiation

• John O' Meara
In summary, the given conversation discusses the proof that if a point P(a,b) is on the rotated ellipse C (x^2 -xy +y^2=4), then so is Q(-a,-b). It also proves that the tangent lines to C at points P and Q are parallel. The equation of the line joining P and Q is y-b=mx-a, where m is the slope of the line and is equal to b/a. The equation of the tangent line is found to be y'=(y-2x)/(2y-x). Replacing x by a and y by b, and then by -a and -b respectively, results in the same slope for both tangent lines.

#### John O' Meara

Prove that if P(a,b) is a point on the rotated ellipse C (whose equation is $$x^2 -xy +y^2=4$$), then so is Q(-a,-b), and that the tangent lines to C at P and Q are parallel.
The equation of the line joining P and Q is
y - b = m(x - a), where $$m= \frac{b-(-b)}{a-(-a)} = \frac{b}{a}$$, then the equation of the line is $$y=\frac{b}{a}x$$. Now, do I substitute for y into the equation for C? To prove the tangent lines are parallel, I have, $$\frac{dy}{dx}=\frac{y-2x}{2y-x}$$.
If I replace x by a/-a and y by b/-b respectively, I get the following for the tangent lines $$\frac{b-2a}{2b-a}, \frac{-b+2a}{-2b+a}$$ which are not the same slope!. I'm studying this on my own so please help. Thanks.

John O' Meara said:
Prove that if P(a,b) is a point on the rotated ellipse C (whose equation is $$x^2 -xy +y^2=4$$), then so is Q(-a,-b), and that the tangent lines to C at P and Q are parallel.
The equation of the line joining P and Q is
y - b = m(x - a), where $$m= \frac{b-(-b)}{a-(-a)} = \frac{b}{a}$$, then the equation of the line is $$y=\frac{b}{a}x$$. Now, do I substitute for y into the equation for C? To prove the tangent lines are parallel, I have, $$\frac{dy}{dx}=\frac{y-2x}{2y-x}$$.
If I replace x by a/-a and y by b/-b respectively, I get the following for the tangent lines $$\frac{b-2a}{2b-a}, \frac{-b+2a}{-2b+a}$$ which are not the same slope!. I'm studying this on my own so please help. Thanks.
I take you have proven that "if P(a,b) is on C then so is Q(-a,-b)". That's simple substitution.
But I have no idea why you are looking at "the line joining P and Q". That has nothing to do with the tangent lines.
Yes, 2x- y- xy'+ 2yy'= 0 so (2y-x)y'= y- 2x and y'= (y- 2x)/(2y-x). If you replace x by a and y by b, you get (b-2a)/(2b-a) as slope of the tangent line. If you replace x by -a and y by -b, you get (-b+2a)/(-2b+ a) as slope- and those are exactly the same: factor a (-1) out of both numerator and denominator in the second fraction.

## 1. What is implicit differentiation?

Implicit differentiation is a mathematical method used to find the derivative of a function that is not expressed explicitly in terms of the independent variable.

## 2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation involves finding the derivative of a function that is expressed explicitly in terms of the independent variable, while implicit differentiation involves finding the derivative of a function that is not explicitly expressed in terms of the independent variable.

## 3. When should I use implicit differentiation?

Implicit differentiation is useful when the dependent variable cannot be easily isolated in the given equation, or when the equation is too complex to be solved explicitly for the dependent variable.

## 4. How do I know when to use implicit differentiation?

If the given equation involves both the dependent and independent variables, and it is difficult or impossible to isolate the dependent variable, then implicit differentiation should be used.

## 5. What are the steps to solve a proof using implicit differentiation?

The steps to solve a proof using implicit differentiation are: 1) differentiate both sides of the equation with respect to the independent variable, 2) simplify the resulting derivative, 3) solve for the derivative of the dependent variable, and 4) substitute the derivative back into the original equation to find the desired value or solve for the dependent variable.