6.7 Q-8

Question Statement

Find the points of intersection of the given conics:

1. $\frac{x^2}{18} + \frac{y^2}{8} = 1$ and $\frac{x^2}{3} - \frac{y^2}{3} = 81$
2. $x^2 + y^2 = 8$ and $x^2 - y^2 = 1$
3. $3x^2 - 4y^2 = 12$ and $3y^2 - 2x^2 = 7$
4. $3x^2 + 5y^2 = 60$ and $9x^2 + y^2 = 124$
5. $\frac{x^2}{9} + \frac{y^2}{9} = 1$ and $\frac{x^2}{2} - \frac{y^2}{4} = 1$

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Background and Explanation

To find the points of intersection between two conics, we need to solve the system of equations that represents these conics. This typically involves:

1. Substituting one equation into the other.
2. Simplifying to find values for the variables.
3. Analyzing the resulting equations to find the points where the conics intersect.

We will use algebraic manipulation and the method of substitution to solve these systems of equations.

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Solution

i. $\frac{x^2}{18} + \frac{y^2}{8} = 1$ and $\frac{x^2}{3} - \frac{y^2}{3} = 81$

1. Simplify and combine the two equations:
   Start by multiplying both equations by appropriate constants to eliminate fractions. From $\frac{x^2}{18} + \frac{y^2}{8} = 1$ and $\frac{x^2}{3} - \frac{y^2}{3} = 81$, we get:

$$8x^2 + 18y^2 = 144 \quad \text{and} \quad x^2 - y^2 = 3$$

Now we have the system:

$$8x^2 + 18y^2 = 144 \quad \text{(1)}$$ $$x^2 - y^2 = 3 \quad \text{(2)}$$

2. Solve the system:
   From equation (2), solve for $x^2$:

$$x^2 = y^2 + 3$$

Substitute this into equation (1):

$$8(y^2 + 3) + 18y^2 = 144$$

Simplifying:

$$8y^2 + 24 + 18y^2 = 144$$ $$26y^2 = 120 \quad \Rightarrow \quad y^2 = \frac{120}{26} = \frac{60}{13}$$

Therefore, $y = \pm \sqrt{\frac{60}{13}}$.

3. Find the corresponding $x$-values:
   Substitute $y^2 = \frac{60}{13}$ into equation (2):

$$x^2 = 3 + \frac{60}{13} = \frac{99}{13}$$

Thus, $x = \pm \sqrt{\frac{99}{13}}$.

ii. $x^2 + y^2 = 8$ and $x^2 - y^2 = 1$

1. Add the equations:
   Adding equation (1) and (2):

$$(x^2 + y^2) + (x^2 - y^2) = 8 + 1$$

Simplifying:

$$2x^2 = 9 \quad \Rightarrow \quad x^2 = \frac{9}{2}$$

So, $x = \pm \frac{3}{\sqrt{2}}$.

2. Substitute $x^2 = \frac{9}{2}$ into equation (1):

$$\frac{9}{2} + y^2 = 8 \quad \Rightarrow \quad y^2 = 8 - \frac{9}{2} = \frac{7}{2}$$

So, $y = \pm \sqrt{\frac{7}{2}}$.

3. Points of intersection:
   The points of intersection are:

$$\left( \pm \frac{3}{\sqrt{2}}, \pm \sqrt{\frac{7}{2}} \right)$$

iii. $3x^2 - 4y^2 = 12$ and $3y^2 - 2x^2 = 7$

1. Multiply equations:
   Multiply equation (1) by 2 and equation (2) by 3, then add the equations:

$$6x^2 - 8y^2 = 24 \quad \text{and} \quad -6x^2 + 9y^2 = 24$$

Adding these gives:

$$y^2 = 45 \quad \Rightarrow \quad y = \pm \sqrt{45}$$

2. Find $x$-values:
   Substitute $y^2 = 45$ into equation (1):

$$3x^2 - 4(45) = 12 \quad \Rightarrow \quad 3x^2 - 180 = 12 \quad \Rightarrow \quad 3x^2 = 192 \quad \Rightarrow \quad x^2 = 64$$

Thus, $x = \pm 8$.

3. Points of intersection:
   The points of intersection are:

$$(\pm 8, \pm \sqrt{45})$$

iv. $3x^2 + 5y^2 = 60$ and $9x^2 + y^2 = 124$

1. Multiply equations:
   Multiply equation (1) by 3 and subtract from equation (2):

$$9x^2 + y^2 = 124 \quad \text{and} \quad 9x^2 + 15y^2 = -180$$

Simplifying:

$$-14y^2 = -56 \quad \Rightarrow \quad y^2 = 4 \quad \Rightarrow \quad y = \pm 2$$

2. Find $x$-values:
   Substitute $y^2 = 4$ into equation (1):

$$3x^2 + 5(4) = 60 \quad \Rightarrow \quad 3x^2 = 40 \quad \Rightarrow \quad x^2 = \frac{40}{3}$$

Thus, $x = \pm \frac{2\sqrt{10}}{3}$.

3. Points of intersection:
   The points of intersection are:

$$\left( \pm \frac{2\sqrt{10}}{3}, \pm 2 \right)$$

v. $\frac{x^2}{9} + \frac{y^2}{9} = 1$ and $\frac{x^2}{2} - \frac{y^2}{4} = 1$

1. Multiply equation (2) by $\frac{1}{2}$ and subtract from equation (1):

$$\frac{x^2}{4} + \frac{y^2}{9} = 1$$

and

$$\frac{x^2}{9} - \frac{y^2}{8} = -\frac{1}{2}$$

Simplifying gives:

$$y^2 = -72$$

This leads to an imaginary solution, so the conics do not intersect.

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Key Formulas or Methods Used

1. Solving Systems of Equations: The primary method used was solving systems of equations by substitution and elimination.

2. Discriminant Condition for Tangency: For the line to be tangent to the conic, the discriminant of the corresponding quadratic equation must be zero.

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Summary of Steps

1. Set up the system of equations for the conics.
2. Use substitution to express one variable in terms of the other.
3. Solve the resulting quadratic equation.
4. Use the solutions to find the points of intersection.
5. If no real solutions exist (i.e., the discriminant is negative), conclude that the conics do not intersect.