Step 1: Understanding the Concept:
Two curves are orthogonal if their tangent lines are perpendicular at the point of intersection. This means the product of their derivatives (slopes) at the intersection point is -1. Step 2: Key Formula or Approach:
1. Find $\frac{dy}{dx}$ (let's call it $m_1$) for the first curve.
2. Find $\frac{dy}{dx}$ (let's call it $m_2$) for the second curve.
3. Use the orthogonality condition $m_1 m_2 = -1$. This gives a relation between $x$ and $y$ at the intersection.
4. Use this relation along with the two original curve equations to solve for the parameter $m$. Step 3: Detailed Explanation:
Curve 1: $x^2 - 4y^2 = 2$
Differentiate implicitly with respect to $x$:
\[ 2x - 8y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{2x}{8y} = \frac{x}{4y} \]
So, slope $m_1 = \frac{x}{4y}$.
Curve 2: $8x^2 + my^2 = 40$
Differentiate implicitly with respect to $x$:
\[ 16x + 2my \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{16x}{2my} = -\frac{8x}{my} \]
So, slope $m_2 = -\frac{8x}{my}$.
Since the curves are orthogonal, $m_1 \cdot m_2 = -1$ at their point of intersection $(x, y)$.
\[ \left(\frac{x}{4y}\right) \left(-\frac{8x}{my}\right) = -1 \]
\[ -\frac{8x^2}{4my^2} = -1 \]
\[ \frac{2x^2}{my^2} = 1 \]
\[ my^2 = 2x^2 \quad \dots \text{ (Equation 3)} \]
Now, we need to find $x^2$ and $y^2$ at the intersection to find $m$.
Substitute $my^2 = 2x^2$ into the equation of Curve 2 ($8x^2 + my^2 = 40$):
\[ 8x^2 + 2x^2 = 40 \]
\[ 10x^2 = 40 \implies x^2 = 4 \]
Substitute $x^2 = 4$ into the equation of Curve 1 ($x^2 - 4y^2 = 2$) to find $y^2$:
\[ 4 - 4y^2 = 2 \]
\[ 4y^2 = 2 \implies y^2 = \frac{1}{2} \]
Finally, substitute the values of $x^2$ and $y^2$ back into Equation 3:
\[ m\left(\frac{1}{2}\right) = 2(4) \]
\[ \frac{m}{2} = 8 \]
\[ m = 16 \]
Step 4: Final Answer:
The value of m is 16.