The equation of the tangent to the curve `y=sqrt(9-2x^(2))` at the point where the ordinate & the abscissa are equal is

Options

A. `2x+y-sqrt3=0`
B. `2x+y-3=0`
C. `2x-y-3sqrt(3)=0`
D. `2x+y-3sqrt(3)=0`

Explanation

To find the equation of the tangent to the curve \( y = \sqrt{9 - 2x^2} \) at the point where the ordinate (y-coordinate) and the abscissa (x-coordinate) are equal, we will follow these steps:

### Step 1: Identify the point where \( x = y \)
Since we need to find the point where the ordinate and abscissa are equal, we set \( y = x \). Thus, we can substitute \( x \) for \( y \) in the equation of the curve:

\[ x = \sqrt{9 - 2x^2} \]

### Step 2: Square both sides to eliminate the square root
Squaring both sides gives us:

\[ x^2 = 9 - 2x^2 \]

### Step 3: Rearrange the equation
Rearranging the equation yields:

\[ x^2 + 2x^2 = 9 \implies 3x^2 = 9 \]

### Step 4: Solve for \( x \)
Dividing both sides by 3 gives:

\[ x^2 = 3 \implies x = \pm \sqrt{3} \]

### Step 5: Find corresponding \( y \) values
Since \( y = x \), we have:

\[ y = \sqrt{3} \quad \text{and} \quad y = -\sqrt{3} \]

Thus, the points where \( x = y \) are \( (\sqrt{3}, \sqrt{3}) \) and \( (-\sqrt{3}, -\sqrt{3}) \).

### Step 6: Find the derivative \( \frac{dy}{dx} \)
To find the slope of the tangent line, we differentiate \( y \):

\[ y = \sqrt{9 - 2x^2} \]

Using the chain rule:

\[ \frac{dy}{dx} = \frac{1}{2\sqrt{9 - 2x^2}} \cdot (-4x) = \frac{-2x}{\sqrt{9 - 2x^2}} \]

### Step 7: Evaluate the derivative at the points
Now we will evaluate the derivative at \( x = \sqrt{3} \):

\[ \frac{dy}{dx} \bigg|_{x = \sqrt{3}} = \frac{-2(\sqrt{3})}{\sqrt{9 - 2(\sqrt{3})^2}} = \frac{-2\sqrt{3}}{\sqrt{9 - 6}} = \frac{-2\sqrt{3}}{\sqrt{3}} = -2 \]

### Step 8: Write the equation of the tangent line
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency:

For the point \( (\sqrt{3}, \sqrt{3}) \):

\[ y - \sqrt{3} = -2(x - \sqrt{3}) \]

Rearranging gives:

\[ y = -2x + 2\sqrt{3} + \sqrt{3} \implies y = -2x + 3\sqrt{3} \]

For the point \( (-\sqrt{3}, -\sqrt{3}) \):

\[ y + \sqrt{3} = -2(x + \sqrt{3}) \]

Rearranging gives:

\[ y = -2x - 2\sqrt{3} - \sqrt{3} \implies y = -2x - 3\sqrt{3} \]

### Final Answer
Thus, the equations of the tangents at the points where the ordinate and abscissa are equal are:

1. \( y = -2x + 3\sqrt{3} \) (at \( (\sqrt{3}, \sqrt{3}) \))
2. \( y = -2x - 3\sqrt{3} \) (at \( (-\sqrt{3}, -\sqrt{3}) \))