# Equation of Tangent and Normal in Different Forms for a Parabola

### Equation of Tangent and Normal in Different Forms for a Parabola:

1. Slope-Intercept Form:

• Tangent Equation: $y=mx+c$ or $x=ny+d$ at the point of tangency.
• The slope $m$ is equal to the derivative of the parabola's equation at the given point.
• The intercept $c$ or $d$ is determined by substituting the coordinates of the point of tangency.

2. Point-Slope Form:

• Tangent Equation: $y-{y}_{1}=m\left(x-{x}_{1}\right)$ or $x-{x}_{1}=m\left(y-{y}_{1}\right)$ at the point of tangency.
• Use the point-slope form with the slope calculated using the derivative of the parabola's equation at the given point.

3. Implicit Form:

• Tangent Equation: $y{y}_{1}=2a\left(x+{x}_{1}\right)$ or $x{x}_{1}=2a\left(y+{y}_{1}\right)$ for ${y}^{2}=4ax$ or ${x}^{2}=4ay$.
• These implicit forms directly relate the coordinates of the point and the slope $a$ of the parabola.

4. Normal Form:

• Normal Equation: $y=-\frac{1}{m}x+e$ or $x=-\frac{1}{n}y+f$ at the point of tangency.
• $m$ or $n$ is the slope of the tangent, and the negative reciprocal gives the slope of the normal.
• $e$ or $f$ are determined by substituting the coordinates of the point of tangency.

5. Parametric Form:

• Parametric Tangent Equation: $y=at+b$ or $x=ct+d$ where $t$ is the parameter.
• Expressing the tangent parametrically helps in visualizing the line's movement as $t$ varies along the curve.

6. Vector Form:

• Tangent Equation: $\stackrel{⃗}{r}=\stackrel{⃗}{a}+\lambda \cdot \stackrel{⃗}{v}$, where $\stackrel{⃗}{r}$ is the position vector and $\stackrel{⃗}{v}$ is the velocity vector.
• Using vector form helps in understanding the direction and movement of the tangent on the parabola.

### Example:

Given: Parabola $y=2{x}^{2}$, and we want to find equations of the tangent and normal at the point $\left(1,2\right)$.

1. Derivation of Tangent and Normal Equations:

a. Slope-Intercept Form:

• Derivative of the Parabola: $\frac{dy}{dx}=4x$

• At $x=1$, $\frac{dy}{dx}$ is $4×1=4$.

• Equation of Tangent: $y-{y}_{1}=m\left(x-{x}_{1}\right)$

• $y-2=4\left(x-1\right)$ simplifies to $y=4x-2$ (Tangent Equation).

• For the Normal, the slope will be the negative reciprocal of the tangent's slope:

• Slope of Normal: $-\frac{1}{4}$

• Equation of Normal: $y-{y}_{1}=m\left(x-{x}_{1}\right)$

• $y-2=-\frac{1}{4}\left(x-1\right)$ simplifies to $y=-\frac{1}{4}x+\frac{9}{4}$ (Normal Equation).

b. Implicit Form:

• For ${y}^{2}=4ax$, the implicit form is $y{y}_{1}=2a\left(x+{x}_{1}\right)$.

• Using $a=\frac{1}{2}$ (from $y=2{x}^{2}$) and $\left({x}_{1},{y}_{1}\right)=\left(1,2\right)$:

• $y×2=2×\frac{1}{2}\left(x+1\right)$ simplifies to $y=x+1$ (Implicit Tangent Equation).

• For the Normal, use the negative reciprocal slope:

• The slope of the Normal is $-\frac{1}{4}$.

• $y{y}_{1}=2a\left(x+{x}_{1}\right)$

• $y×2=2×\frac{1}{2}\left(x+1\right)$ simplifies to $y=x+1$ (Implicit Normal Equation).

c. Parametric Form:

• The parametric equation for the tangent is $y=at+b$ or $x=ct+d$.

• At $x=1$, $y=2{x}^{2}=2\left(1{\right)}^{2}=2$.

• $y=2t+0$ and $x=t+1$ (Parametric Tangent Equation).

• For the Normal, the slope will be the negative reciprocal of the tangent's slope:

• The slope of the Normal is $-\frac{1}{4}$.

• $y=-\frac{1}{4}t+\frac{9}{4}$ and $x=t+1$ (Parametric Normal Equation).

### Analysis:

• The equations for the tangent and normal to the parabola $y=2{x}^{2}$ at the point $\left(1,2\right)$ are obtained in different forms.
• Each form emphasizes various aspects, such as slope-intercept, implicit, and parametric representations, offering unique insights into the behavior of the curves at that point.