The slope of the tangent to the curve `2x^(2) + 3y^(2) =5` at the point whose abscissa is -2, is

Options

A. `-2/3`
B. `2/3`
C. `3/2`
D. `-3/2`

Answer

To find the slope of the tangent to the curve \(2x^2 + 3y^2 = 5\) at the point where the abscissa (x-coordinate) is -2, we will follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation of the curve:
\[ 2x^2 + 3y^2 = 5 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(2x^2) + \frac{d}{dx}(3y^2) = \frac{d}{dx}(5) \]
This gives us:
\[ 4x + 6y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ 6y \frac{dy}{dx} = -4x \]
\[ \frac{dy}{dx} = -\frac{4x}{6y} = -\frac{2x}{3y} \]

### Step 3: Find the corresponding \(y\) value when \(x = -2\)
Substituting \(x = -2\) into the original equation to find \(y\):
\[ 2(-2)^2 + 3y^2 = 5 \]
\[ 2(4) + 3y^2 = 5 \]
\[ 8 + 3y^2 = 5 \]
\[ 3y^2 = 5 - 8 \]
\[ 3y^2 = -3 \]
\[ y^2 = -1 \]

### Step 4: Analyze the result
Since \(y^2 = -1\) does not yield a real number solution, this indicates that there is no real \(y\) value corresponding to \(x = -2\). Therefore, the slope of the tangent at this point is not defined.

### Final Answer
The slope of the tangent to the curve \(2x^2 + 3y^2 = 5\) at the point where the abscissa is -2 is **not defined**.

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