The graph of the equation x2+xy+y2=5 is an ellipse lying obliquely in the plane, as illustrated in the figure below.
 
a.  Compute dydx. 

dydx= . 


b.  The ellipse has two horizontal tangents. Find an equation of the upper one. 

The upper horizontal tangent line is defined by the equation y= . 


c.  The ellipse has two vertical tangents. Find an equation of the rightmost one. 

The rightmost vertical tangent line is defined by the equation x= . 


d.  Find the point at which the rightmost vertical tangent line touches the ellipse. 

The rightmost vertical tangent line touches the ellipse at the point . 


Hint: The horizontal tangent is of course characterized by dydx=0. To find the vertical tangent use symmetry, or solve dxdy=0.


Let's go through each part of this problem systematically.

### Part (a): Compute \(\frac{dy}{dx}\).

Given:
\[
x^2 + xy + y^2 = 5
\]

To find \(\frac{dy}{dx}\), we implicitly differentiate both sides with respect to \(x\).

1. Differentiate \(x^2\):
   \[
   \frac{d}{dx}(x^2) = 2x
   \]

2. Differentiate \(xy\) using the product rule:
   \[
   \frac{d}{dx}(xy) = y + x \frac{dy}{dx}
   \]

3. Differentiate \(y^2\) using the chain rule:
   \[
   \frac{d}{dx}(y^2) = 2y \frac{dy}{dx}
   \]

So, the differentiated equation becomes:
\[
2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0
\]

Now, we can solve for \(\frac{dy}{dx}\):
\[
x \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - y
\]
\[
\frac{dy}{dx}(x + 2y) = -2x - y
\]
\[
\frac{dy}{dx} = \frac{-2x - y}{x + 2y}
\]

So, 
\[
\frac{dy}{dx} = \frac{-2x - y}{x + 2y}
\]

### Part (b): Find the equation of the upper horizontal tangent.

For a horizontal tangent, we need \(\frac{dy}{dx} = 0\).

Setting the numerator of \(\frac{dy}{dx}\) to zero:
\[
-2x - y = 0
\]
\[
y = -2x
\]

Substitute \(y = -2x\) into the original equation:
\[
x^2 + x(-2x) + (-2x)^2 = 5
\]
\[
x^2 - 2x^2 + 4x^2 = 5
\]
\[
3x^2 = 5
\]
\[
x^2 = \frac{5}{3}
\]
\[
x = \pm \sqrt{\frac{5}{3}}
\]

For the **upper tangent**, we choose the positive \(x\)-value:
\[
x = \sqrt{\frac{5}{3}}
\]

Then,
\[
y = -2\sqrt{\frac{5}{3}}
\]

Thus, the equation of the upper horizontal tangent line is:
\[
y = -2\sqrt{\frac{5}{3}}
\]

### Part (c): Find the equation of the rightmost vertical tangent.

For a vertical tangent, we need \(\frac{dx}{dy} = 0\), which means \(\frac{dy}{dx}\) has an undefined denominator.

Set the denominator of \(\frac{dy}{dx}\) to zero:
\[
x + 2y = 0
\]
\[
x = -2y
\]

Substitute \(x = -2y\) into the original equation:
\[
(-2y)^2 + (-2y)y + y^2 = 5
\]
\[
4y^2 - 2y^2 + y^2 = 5
\]
\[
3y^2 = 5
\]
\[
y^2 = \frac{5}{3}
\]
\[
y = \pm \sqrt{\frac{5}{3}}
\]

For the **rightmost tangent**, we take the positive \(y\)-value:
\[
y = \sqrt{\frac{5}{3}}
\]

Then,
\[
x = -2\sqrt{\frac{5}{3}}
\]

Thus, the equation of the rightmost vertical tangent line is:
\[
x = -2\sqrt{\frac{5}{3}}
\]

### Part (d): Find the point where the rightmost vertical tangent touches the ellipse.

The point where the rightmost vertical tangent touches the ellipse is:
\[
\left(-2\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right)
\]

---

### Summary of Answers
- **(a)** \(\frac{dy}{dx} = \frac{-2x - y}{x + 2y}\)
- **(b)** The equation of the upper horizontal tangent line is \(y = -2\sqrt{\frac{5}{3}}\)
- **(c)** The equation of the rightmost vertical tangent line is \(x = -2\sqrt{\frac{5}{3}}\)
- **(d)** The point where the rightmost vertical tangent touches the ellipse is \(\left(-2\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right)\)

Do you have any questions on these solutions or want additional clarification? 

Here are some related questions for further exploration:
1. How would the solution change if the equation was \(x^2 + xy + y^2 = c\) with \(c \neq 5\)?
2. What is the significance of implicit differentiation in finding slopes of tangent lines in conic sections?
3. Can we parametrize this ellipse and find tangent lines without differentiation?
4. How can we identify if other conic sections (like hyperbolas) have horizontal or vertical tangents?
5. What are the general methods for finding tangent lines to ellipses in different orientations?

**Tip**: Implicit differentiation is invaluable when dealing with non-standard forms of conics, as it allows direct computation of derivatives without explicitly solving for \(y\) in terms of \(x\).

The graph of the equation x^2 + xy + y^2 = 5 is an ellipse lying obliquely in the plane, as illustrated in the figure below.

a. Compute dy/dx.

dy/dx = . 

b. The ellipse has two horizontal tangents. Find an equation of the upper one.

The upper horizontal tangent line is defined by the equation y = . 

c. The ellipse has two vertical tangents. Find an equation of the rightmost one.

The rightmost vertical tangent line is defined by the equation x = . 

d. Find the point at which the rightmost vertical tangent line touches the ellipse.

The rightmost vertical tangent line touches the ellipse at the point .

Explore how to find the equations of horizontal and vertical tangent lines to the ellipse defined by x^2 + xy + y^2 = 5 using implicit differentiation. This detailed solution covers steps for calculating dy/dx, identifying tangent lines, and finding points of tangency, suitable for Grades 11-12 students.

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