Slope of variable line which passes through (1,2)

Such that the area bounded by this line and

y^2_8x_4y+12=0is 4/3sq.unit is ..

To find the slope of the variable line that passes through the point (1, 2) and creates an area of 4/3 square units with the curve defined by the equation \(y^2 - 8x - 4y + 12 = 0\), we first need to analyze the given curve and then determine how the line interacts with it.

Understanding the Curve

The equation \(y^2 - 8x - 4y + 12 = 0\) can be rearranged to make it easier to work with. Let's rewrite it:

• Rearranging gives us \(y^2 - 4y + 12 = 8x\).
• Completing the square for the \(y\) terms: \(y^2 - 4y + 4 = 8x - 8\).
• This simplifies to \((y - 2)^2 = 8(x - 1)\).

This is a parabola that opens to the right, with its vertex at the point (1, 2).

Setting Up the Line

Next, we consider a line that passes through the point (1, 2). The equation of a line in slope-intercept form is given by:

y - 2 = m(x - 1)

Here, \(m\) represents the slope of the line. Rearranging this gives us:

y = mx - m + 2

Finding the Area

The area between the line and the parabola can be calculated using integration. To find the points of intersection, we set the equations equal to each other:

mx - m + 2 = (y - 2)^2 + 8

Substituting \(y\) from the line's equation into the parabola's equation will yield a quadratic equation in terms of \(x\). The area between the curves can be calculated by integrating the difference of the functions over the interval defined by the intersection points.

Calculating the Area

The area \(A\) between the curves can be expressed as:

A = ∫[x1 to x2] (parabola - line) dx

Given that the area is specified as \(4/3\) square units, we can set up the equation:

∫[x1 to x2] ((y from parabola) - (y from line)) dx = 4/3

Finding the Slope

To find the slope \(m\), we need to solve the area equation for different values of \(m\). This typically involves substituting \(m\) into the area equation and solving for \(m\) such that the area equals \(4/3\). This may require numerical methods or graphing techniques to find the appropriate slope values.

In practice, you would calculate the definite integral and solve for \(m\) using algebraic techniques or numerical approximation methods, depending on the complexity of the resulting equations.

Conclusion

In summary, the slope \(m\) of the variable line that passes through the point (1, 2) and bounds an area of \(4/3\) square units with the given parabola can be determined through integration and solving for \(m\) in the area equation. This process involves understanding the relationship between the line and the curve, as well as applying calculus to find the area between them.