Equation of the tangent to the curve `2x^(2)+3y^(2)-5=0` at (1, 1) is

Options

A. `2x-3y-5=0`
B. `2x+3y-5=0`
C. `2x+3y+5=0`
D. `3x+2y+5=0`.

Answer

To find the equation of the tangent to the curve \(2x^2 + 3y^2 - 5 = 0\) at the point \((1, 1)\), we will follow these steps:

### Step 1: Differentiate the equation of the curve
We start with the equation of the curve:
\[ 2x^2 + 3y^2 - 5 = 0 \]
We differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(2x^2) + \frac{d}{dx}(3y^2) - \frac{d}{dx}(5) = 0 \]
Using the chain rule on \(3y^2\), we get:
\[ 4x + 6y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ 6y \frac{dy}{dx} = -4x \]
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = -\frac{4x}{6y} = -\frac{2x}{3y} \]

### Step 3: Find the slope at the point (1, 1)
Now we substitute the point \((1, 1)\) into the derivative to find the slope of the tangent line at that point:
\[ \frac{dy}{dx} \bigg|_{(1,1)} = -\frac{2(1)}{3(1)} = -\frac{2}{3} \]

### Step 4: Use point-slope form to find the equation of the tangent
The point-slope form of the equation of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = -\frac{2}{3}\), \(x_1 = 1\), and \(y_1 = 1\):
\[ y - 1 = -\frac{2}{3}(x - 1) \]

### Step 5: Simplify the equation
Now we simplify this equation:
\[ y - 1 = -\frac{2}{3}x + \frac{2}{3} \]
Adding 1 to both sides:
\[ y = -\frac{2}{3}x + \frac{2}{3} + 1 \]
Converting 1 to a fraction with a denominator of 3:
\[ y = -\frac{2}{3}x + \frac{2}{3} + \frac{3}{3} = -\frac{2}{3}x + \frac{5}{3} \]

### Step 6: Rearranging to standard form
To convert this into standard form \(Ax + By + C = 0\):
\[ \frac{2}{3}x + y - \frac{5}{3} = 0 \]
Multiplying through by 3 to eliminate the fractions:
\[ 2x + 3y - 5 = 0 \]

### Final Answer
Thus, the equation of the tangent to the curve at the point \((1, 1)\) is:
\[ 2x + 3y - 5 = 0 \]