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: yy1=m(xx1) or xx1=m(yy1) 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: yy1=2a(x+x1) or xx1=2a(y+y1) for y2=4ax or x2=4ay.
    • These implicit forms directly relate the coordinates of the point and the slope a of the parabola.

4. Normal Form:

  • Normal Equation: y=1mx+e or x=1ny+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: r=a+λv, where r is the position vector and 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=2x2, and we want to find equations of the tangent and normal at the point (1,2).

1. Derivation of Tangent and Normal Equations:

a. Slope-Intercept Form:

  • Derivative of the Parabola: dydx=4x

  • At x=1, dydx is 4×1=4.

  • Equation of Tangent: yy1=m(xx1)

  • 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: 14

  • Equation of Normal: yy1=m(xx1)

  • y2=14(x1) simplifies to y=14x+94 (Normal Equation).

b. Implicit Form:

  • For y2=4ax, the implicit form is yy1=2a(x+x1).

  • Using a=12 (from y=2x2) and (x1,y1)=(1,2):

  • y×2=2×12(x+1) simplifies to y=x+1 (Implicit Tangent Equation).

  • For the Normal, use the negative reciprocal slope:

  • The slope of the Normal is 14.

  • yy1=2a(x+x1)

  • y×2=2×12(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=2x2=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 14.

  • y=14t+94 and x=t+1 (Parametric Normal Equation).

Analysis:

  • The equations for the tangent and normal to the parabola y=2x2 at the point (1,2) 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.