4.5 Q-9

Question Statement

Find the area of the region bounded by the curves:

1. $10x^2 - xy - 21y^2 = 0$
2. $x + y + 1 = 0$

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Background and Explanation

To solve this problem, we need to:

1. Understand the geometry of the given equations. The first equation represents a conic section (hyperbola), and the second is a straight line.
2. Identify the points of intersection of the line with the hyperbola by solving the equations simultaneously.
3. Use these points to compute the area of the triangle formed by the intersecting points and the origin.

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Solution

Step 1: Rewrite the Conic Equation

The given conic equation is:
$10x^2 - xy - 21y^2 = 0$

We rewrite it as:
$\frac{-21y^2}{-21x^2} + \frac{1}{21}\left(\frac{y}{x}\right) - \frac{10}{21} = 0$

This is a quadratic equation in $\frac{y}{x}$:
$\frac{y}{x} = \frac{\frac{1}{21} \pm \sqrt{\left(\frac{1}{21}\right)^2 - 4(1)\left(-\frac{10}{21}\right)}}{2}$

Step 2: Solve for $\frac{y}{x}$

Simplifying:
$\frac{y}{x} = \frac{\frac{1}{21} \pm \sqrt{\frac{1}{441} + \frac{40}{21}}}{2}$
$\frac{y}{x} = \frac{\frac{1}{21} \pm \sqrt{\frac{841}{441}}}{2}$
$\frac{y}{x} = \frac{\frac{1}{21} \pm \frac{29}{21}}{2}$

We get two slopes:

1. $\frac{y}{x} = \frac{-1 + 29}{42} = \frac{28}{42} = \frac{2}{3} \quad \rightarrow \quad 2x - 3y = 0$
2. $\frac{y}{x} = \frac{-1 - 29}{42} = \frac{-30}{42} = \frac{-5}{7} \quad \rightarrow \quad 5x + 7y = 0$

Step 3: Find Points of Intersection

The straight line $x + y + 1 = 0$ can be rewritten as:
$x = -y - 1$

Solve $x + y + 1 = 0$ and $2x - 3y = 0$:

Substitute $x = -y - 1$ into $2x - 3y = 0$:
$2(-y - 1) - 3y = 0$
$-2y - 2 - 3y = 0 \quad \Rightarrow \quad -5y = 2 \quad \Rightarrow \quad y = -\frac{2}{5}$

Substitute $y = -\frac{2}{5}$ into $x = -y - 1$:
$x = -\left(-\frac{2}{5}\right) - 1 = \frac{2}{5} - 1 = -\frac{3}{5}$

Point of intersection:
$A\left(-\frac{3}{5}, -\frac{2}{5}\right)$

Solve $x + y + 1 = 0$ and $5x + 7y = 0$:

Substitute $x = -y - 1$ into $5x + 7y = 0$:
$5(-y - 1) + 7y = 0$
$-5y - 5 + 7y = 0 \quad \Rightarrow \quad 2y = 5 \quad \Rightarrow \quad y = \frac{5}{2}$

Substitute $y = \frac{5}{2}$ into $x = -y - 1$:
$x = -\frac{5}{2} - 1 = -\frac{7}{2}$

Point of intersection:
$B\left(-\frac{7}{2}, \frac{5}{2}\right)$

Origin as a Third Point:

The lines $2x - 3y = 0$ and $5x + 7y = 0$ intersect at the origin:
$O(0, 0)$

Step 4: Compute the Area of Triangle $\triangle OAB$

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:
$\Delta = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substitute $O(0, 0)$, $A\left(-\frac{3}{5}, -\frac{2}{5}\right)$, and $B\left(-\frac{7}{2}, \frac{5}{2}\right)$:
$\Delta = \frac{1}{2} \left| 0 + \left(-\frac{3}{5}\right)\left(\frac{5}{2} - 0\right) + \left(-\frac{7}{2}\right)\left(0 - \left(-\frac{2}{5}\right)\right) \right|$
$\Delta = \frac{1}{2} \left| 0 - \frac{15}{10} - \frac{14}{10} \right|$
$\Delta = \frac{1}{2} \left| -\frac{29}{10} \right| = \frac{29}{20} , \text{square units}$

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Key Formulas or Methods Used

1. Quadratic equations to solve for slopes of the lines from the conic.
2. Simultaneous equations to find points of intersection.
3. Triangle area formula using determinant.

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Summary of Steps

1. Rewrite the conic equation to find its slopes, forming two lines.
2. Solve the line equations to find their intersection points with $x + y + 1 = 0$.
3. Use the origin and intersection points to calculate the area of the triangle.
4. Final result: $\frac{29}{20}$ square units.