show that the family of parabola x2=4a(y+a) is self orthogonal

To show that the family of parabolas \( x^2 = 4a(y + a) \) is self-orthogonal, we need to demonstrate that at the points where two parabolas from this family intersect, their tangents are perpendicular.

### Steps to Prove Self-Orthogonality

1. **Equation of the Parabola:**
   \[
   x^2 = 4a(y + a)
   \]
   
2. **Finding the Slope of the Tangent:**

   To find the slope of the tangent at any point on the parabola, differentiate the equation implicitly with respect to \( x \):

   \[
   \frac{d}{dx} \left( x^2 \right) = \frac{d}{dx} \left( 4a(y + a) \right)
   \]

   \[
   2x = 4a \frac{dy}{dx}
   \]

   \[
   \frac{dy}{dx} = \frac{x}{2a}
   \]

   Hence, the slope of the tangent line at any point \((x, y)\) on the parabola is:

   \[
   m = \frac{x}{2a}
   \]

3. **Orthogonality Condition:**

   Two curves are orthogonal at their intersection if the product of their slopes at the intersection points is \(-1\).

4. **Intersection Points of Two Parabolas:**

   Consider two parabolas from the family:
   
   \[
   x^2 = 4a_1(y + a_1) \quad \text{and} \quad x^2 = 4a_2(y + a_2)
   \]

   Setting the two equations equal to find their intersection points:

   \[
   4a_1(y + a_1) = 4a_2(y + a_2)
   \]

   \[
   a_1(y + a_1) = a_2(y + a_2)
   \]

   \[
   a_1 y + a_1^2 = a_2 y + a_2^2
   \]

   \[
   y(a_1 - a_2) = a_2^2 - a_1^2
   \]

   \[
   y = \frac{a_2^2 - a_1^2}{a_1 - a_2} = \frac{(a_2 - a_1)(a_2 + a_1)}{a_1 - a_2} = -(a_1 + a_2)
   \]

   So, the intersection points occur at \( y = -(a_1 + a_2) \).

5. **Slopes at the Intersection Points:**

   For \( a_1 \):

   \[
   m_1 = \frac{x}{2a_1}
   \]

   For \( a_2 \):

   \[
   m_2 = \frac{x}{2a_2}
   \]

   The orthogonality condition requires:

   \[
   m_1 \cdot m_2 = \left( \frac{x}{2a_1} \right) \left( \frac{x}{2a_2} \right) = -1
   \]

   Simplifying:

   \[
   \frac{x^2}{4a_1a_2} = -1
   \]

   From the original parabola equation, \( x^2 = 4a(y + a) \), substituting \( y = -(a_1 + a_2) \):

   \[
   x^2 = 4a_1(-(a_1 + a_2) + a_1) = 4a_1(-

Learn how to prove the self-orthogonality of the family of parabolas x^2 = 4a(y + a) using differential calculus and the concept of orthogonality of curves. This detailed mathematical problem explores the intersection points and slopes of tangents, demonstrating perpendicularity.

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