The curve $$ 12 x^{2}-5 y^{2}=7 $$ intersects the line $$ 2 p^{2} x-5 y=7 $$ at the point $$ (1, p) $$. (i) Find the value of $$ p $$. $$ x, y $$ (ii) Find the coordinates of the other point of intersection.

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To solve the problem, we'll address each part step-by-step.

### Given:
- **Curve:** $$ 12x^{2} - 5y^{2} = 7 $$
- **Line:** $$ 2p^{2}x - 5y = 7 $$
- **Intersection Point:** $$ (1, p) $$

### Part (i): Find the value of $$ p $$

Since $$ (1, p) $$ lies on both the curve and the line, we'll substitute $$ x = 1 $$ and $$ y = p $$ into both equations.

1. **Substitute into the curve:**
   $$
   12(1)^2 - 5p^2 = 7 \
   12 - 5p^2 = 7 \
   -5p^2 = -5 \
   p^2 = 1 \
   p = \pm 1
   $$
   
2. **Substitute into the line with $$ p = \pm 1 $$:**
   $$
   2p^2(1) - 5p = 7 \
   2(p^2) - 5p - 7 = 0
   $$
   
   Since $$ p^2 = 1 $$:
   $$
   2(1) - 5p - 7 = 0 \
   2 - 5p - 7 = 0 \
   -5p - 5 = 0 \
   -5p = 5 \
   p = -1
   $$
   
   **Thus, $$ p = -1 $$.**

### Part (ii): Find the coordinates of the other point of intersection

Now that we know $$ p = -1 $$, we can find the other intersection point.

1. **Equation of the line with $$ p = -1 $$:**
   $$
   2(-1)^2x - 5y = 7 \
   2x - 5y = 7
   $$
   
2. **Express $$ y $$ in terms of $$ x $$:**
   $$
   2x - 5y = 7 \
   5y = 2x - 7 \
   y = \frac{2x - 7}{5}
   $$
   
3. **Substitute $$ y $$ into the curve equation:**
   $$
   12x^2 - 5\left(\frac{2x - 7}{5}\right)^2 = 7 \
   12x^2 - \frac{(4x^2 - 28x + 49)}{5} = 7 \
   Multiply all terms by 5 to eliminate the denominator:
   \[
   60x^2 - (4x^2 - 28x + 49) = 35 \
   56x^2 + 28x - 84 = 0 \
   Divide by 28:
   \[
   2x^2 + x - 3 = 0
   $$
   
4. **Solve the quadratic equation:**
   $$
   x = \frac{-1 \pm \sqrt{1 + 24}}{4} \
   x = \frac{-1 \pm 5}{4}
   $$
   Thus, $$ x = 1 $$ or $$ x = -\frac{3}{2} $$.

5. **Find corresponding $$ y $$ values:**
   - For $$ x = 1 $$:
     $$
     y = \frac{2(1) - 7}{5} = \frac{-5}{5} = -1
     $$
     (This is the given point.)
     
   - For $$ x = -\frac{3}{2} $$:
     $$
     y = \frac{2(-\frac{3}{2}) - 7}{5} = \frac{-3 - 7}{5} = \frac{-10}{5} = -2
     $$
     
   **Thus, the other point of intersection is $$ \left(-\frac{3}{2}, -2\right) $$.**

### **Final Answer:**

(i) The value of $$ p $$ is $$ -1 $$.

(ii) The other point of intersection is $$ \left(-\dfrac{3}{2},\; -2\right) $$.
$$ p = -1 $$ and the other intersection point is $$ \left(-\frac{3}{2}, -2\right) $$.

The curve \( 12 x^{2}-5 y^{2}=7 \) intersects the line \( 2 p^{2} x-5 y=7 \) at the point \( (1, p) \). (i) Find the value of \( p \). \( x, y \) (ii) Find the coordinates of the other point of intersection.

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