Equation of Tangent and Normal in Different Forms for a Parabola
Equation of Tangent and Normal in Different Forms for a Parabola:
1. SlopeIntercept 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. PointSlope Form:
 Tangent Equation: $y{y}_{1}=m(x{x}_{1})$ or $x{x}_{1}=m(y{y}_{1})$ at the point of tangency.
 Use the pointslope form with the slope calculated using the derivative of the parabola's equation at the given point.
3. Implicit Form:
 Tangent Equation: $y{y}_{1}=2a(x+{x}_{1})$ or $x{x}_{1}=2a(y+{y}_{1})$ 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: $\stackrel{\u20d7}{r}=\stackrel{\u20d7}{a}+\lambda \cdot \stackrel{\u20d7}{v}$, where $\stackrel{\u20d7}{r}$ is the position vector and $\stackrel{\u20d7}{v}$ 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 $(1,2)$.
1. Derivation of Tangent and Normal Equations:
a. SlopeIntercept Form:

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

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

Equation of Tangent: $y{y}_{1}=m(x{x}_{1})$

$y2=4(x1)$ simplifies to $y=4x2$ (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: $y{y}_{1}=m(x{x}_{1})$

$y2=\frac{1}{4}(x1)$ simplifies to $y=\frac{1}{4}x+\frac{9}{4}$ (Normal Equation).
b. Implicit Form:

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

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

$y\times 2=2\times \frac{1}{2}(x+1)$ simplifies to $y=x+1$ (Implicit Tangent Equation).

For the Normal, use the negative reciprocal slope:

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

$y{y}_{1}=2a(x+{x}_{1})$

$y\times 2=2\times \frac{1}{2}(x+1)$ 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 $x=1$, $y=2{x}^{2}=2(1{)}^{2}=2$.

$y=2t+0$ 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 $(1,2)$ are obtained in different forms.
 Each form emphasizes various aspects, such as slopeintercept, implicit, and parametric representations, offering unique insights into the behavior of the curves at that point.