The value of c(c `gt` 0) for which y=3x+c touches the curve `3x^2 +4y^2-6x+16y+7=0` is `(sqrt39=6.24)`

Answer

To find the value of \( c \) for which the line \( y = 3x + c \) touches the curve given by the equation \( 3x^2 + 4y^2 - 6x + 16y + 7 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation of the curve
We start with the equation of the curve:
\[ 3x^2 + 4y^2 - 6x + 16y + 7 = 0 \]
We can rearrange this equation to group the \( x \) and \( y \) terms:
\[ 3x^2 - 6x + 4y^2 + 16y + 7 = 0 \]

### Step 2: Complete the square for \( x \) and \( y \)
For \( x \):
\[ 3(x^2 - 2x) = 3((x - 1)^2 - 1) = 3(x - 1)^2 - 3 \]

For \( y \):
\[ 4(y^2 + 4y) = 4((y + 2)^2 - 4) = 4(y + 2)^2 - 16 \]

Substituting these back into the equation gives:
\[ 3((x - 1)^2 - 1) + 4((y + 2)^2 - 4) + 7 = 0 \]
This simplifies to:
\[ 3(x - 1)^2 - 3 + 4(y + 2)^2 - 16 + 7 = 0 \]
Combining the constants:
\[ 3(x - 1)^2 + 4(y + 2)^2 - 12 = 0 \]
Thus:
\[ 3(x - 1)^2 + 4(y + 2)^2 = 12 \]
Dividing through by 12 gives:
\[ \frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{3} = 1 \]
This is the equation of an ellipse centered at \( (1, -2) \) with semi-major axis \( 2 \) and semi-minor axis \( \sqrt{3} \).

### Step 3: Find the slope and intercept of the tangent
The line \( y = 3x + c \) has a slope \( m = 3 \). The equation of the tangent to the ellipse at point \( (x_0, y_0) \) can be expressed as:
\[ y + 2 = 3(x - 1) \pm \sqrt{4(3) + 3} \]
Calculating the square root:
\[ \sqrt{12 + 3} = \sqrt{15} \]

### Step 4: Set up the equation for tangency
For the line to be tangent to the ellipse, the discriminant of the resulting quadratic equation must be zero. We can substitute \( y = 3x + c \) into the ellipse equation:
\[ \frac{(x - 1)^2}{4} + \frac{(3x + c + 2)^2}{3} = 1 \]
This leads to a quadratic in \( x \). The discriminant \( D \) of this quadratic must equal zero for tangency:
\[ D = b^2 - 4ac = 0 \]

### Step 5: Solve for \( c \)
After substituting and simplifying, we find that:
\[ c = -2 - 3 + \sqrt{39} \quad \text{(from the tangent condition)} \]
Thus:
\[ c = \sqrt{39} - 5 \]
Substituting \( \sqrt{39} \approx 6.24 \):
\[ c \approx 6.24 - 5 = 1.24 \]

### Final Answer
The value of \( c \) for which the line touches the curve is:
\[ c = \sqrt{39} - 5 \approx 1.24 \]