Equation of tangent at(t,t^2/2)is
y=tx-t^2/2 ⇒ t=dy/dx
### Differential equation is
(dy/dx)^2-2x dy/dx+2y=0


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The equation of the tangent at the point \((t, \frac{t^2}{2})\) is given by:

\[ y = tx - \frac{t^2}{2} \]

From this, it follows that:

\[ t = \frac{dy}{dx} \]

The corresponding differential equation can be expressed as:

\[ \left(\frac{dy}{dx}\right)^2 - 2x \frac{dy}{dx} + 2y = 0 \]

Would you like more details or have any questions?

Here are some related questions:
1. How can you derive the equation of a tangent at a given point on a curve?
2. What are the general steps to find the slope of a tangent line?
3. How can the differential equation \(\left(\frac{dy}{dx}\right)^2 - 2x \frac{dy}{dx} + 2y = 0\) be solved?
4. What does it mean when we say \( t = \frac{dy}{dx} \) in this context?
5. How can we interpret the solution to the differential equation geometrically?
6. How would the equation of the tangent change if the curve were different?
7. What are the possible solutions to the quadratic differential equation provided?
8. How can such differential equations be applied in physics or engineering?

**Tip:** When deriving differential equations from geometric conditions like tangency, remember that the slope of the tangent line is crucial.

Explore the equation of tangent at (t, t^2/2) and its corresponding differential equation solution. Understand how to derive and apply the tangent line equation and solve the associated quadratic differential equation. Suitable for advanced high school students and beyond.

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