If `(x,y)` lies on the ellipse `x^(2)+2y^(3) = 2`, then maximum value of `x^(2)+y^(2)+ sqrt(2)xy - 1` is

Options

A. `(sqrt(5)+1)/(2)`
B. `(sqrt(5)-1)/(2)`
C. `(sqrt(5)+1)/(4)`
D. `(sqrt(5)-1)/(4)`

Explanation

To find the maximum value of the expression \( x^2 + y^2 + \sqrt{2}xy - 1 \) given that \( (x, y) \) lies on the ellipse defined by the equation \( x^2 + 2y^2 = 2 \), we can follow these steps:

### Step 1: Parametrize the ellipse
The equation of the ellipse can be rewritten in a parametric form. We can let:
\[ x = \sqrt{2} \cos \theta, \quad y = \sqrt{1} \sin \theta \]
This satisfies the ellipse equation since:
\[ x^2 + 2y^2 = (\sqrt{2} \cos \theta)^2 + 2(\sin \theta)^2 = 2 \cos^2 \theta + 2 \sin^2 \theta = 2(\cos^2 \theta + \sin^2 \theta) = 2 \]

### Step 2: Substitute into the expression
Now, substitute \( x \) and \( y \) into the expression \( x^2 + y^2 + \sqrt{2}xy - 1 \):
\[ x^2 = 2 \cos^2 \theta, \quad y^2 = \sin^2 \theta, \quad xy = \sqrt{2} \cos \theta \sin \theta \]
Thus, we have:
\[ x^2 + y^2 + \sqrt{2}xy - 1 = 2 \cos^2 \theta + \sin^2 \theta + \sqrt{2}(\sqrt{2} \cos \theta \sin \theta) - 1 \]
\[ = 2 \cos^2 \theta + \sin^2 \theta + 2 \cos \theta \sin \theta - 1 \]

### Step 3: Simplify the expression
Now, simplify the expression:
\[ = 2 \cos^2 \theta + \sin^2 \theta + 2 \cos \theta \sin \theta - 1 \]
Combine \( 2 \cos^2 \theta + \sin^2 \theta \):
\[ = (2 \cos^2 \theta + \sin^2 \theta) + 2 \cos \theta \sin \theta - 1 \]
\[ = (2 \cos^2 \theta + \sin^2 \theta - 1) + 2 \cos \theta \sin \theta \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = (2 \cos^2 \theta - 1 + \sin^2 \theta) + 2 \cos \theta \sin \theta \]
\[ = 2 \cos^2 \theta - 1 + 1 - \cos^2 \theta + 2 \cos \theta \sin \theta \]
\[ = \cos^2 \theta + 2 \cos \theta \sin \theta \]

### Step 4: Use trigonometric identities
Now, we can express this in terms of a single angle:
\[ = \cos^2 \theta + \sin 2\theta \]
Using \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \):
\[ = \frac{1 + \cos 2\theta}{2} + \sin 2\theta \]
\[ = \frac{1}{2} + \frac{\cos 2\theta}{2} + \sin 2\theta \]

### Step 5: Find the maximum value
To find the maximum value of \( \frac{1}{2} + \frac{\cos 2\theta}{2} + \sin 2\theta \), we can rewrite it as:
\[ = \frac{1}{2} + \frac{1}{2}(\cos 2\theta + 2\sin 2\theta) \]
The maximum value of \( \cos 2\theta + 2\sin 2\theta \) can be found using the formula for the maximum of \( a \cos x + b \sin x \):
\[ \sqrt{1^2 + 2^2} = \sqrt{5} \]
Thus, the maximum value of \( \frac{1}{2} + \frac{1}{2} \sqrt{5} \) is:
\[ \frac{1 + \sqrt{5}}{2} \]

### Final Answer
The maximum value of \( x^2 + y^2 + \sqrt{2}xy - 1 \) is:
\[ \frac{1 + \sqrt{5}}{2} \]