Equation of Tangent and Normal to Parabola at a Specific Point
Understanding the equations of the tangent and normal to a parabola at a specific point is crucial in various areas of mathematics, including calculus and geometry. This article will guide you through the process of finding these equations for the parabola given by the equation x^2 16a at the point (-4a, 0).
Introduction to Parabolas
A parabola is a conic section defined as the set of all points in a plane that are equidistant from a fixed point (focus) and a fixed straight line (directrix). The standard form of the equation for a parabola that opens along the x-axis is given by:
Equation: x^2 4ay
For the given problem, the equation of the parabola is x^2 16a. Notice that this can be rewritten as:
Required form: x^2 4(4a)y
This represents a parabola that opens to the right, with the focus at (4a, 0) and the directrix x -4a.
Finding the Point on the Parabola
The problem mentions the point with coordinate -4a. Given the equation of the parabola, we see that the y-coordinate of the point is zero because:
Equation: x^2 16a
When x -4a, we have:
-4a)^2 16a
-16a^2 16a
This equation does not have a real solution because:
16a^2 16a 0
16a(a 1) 0
Which implies:
a 0
or
a -1
However, since a is a parameter for the parabola, it cannot be zero because it would make the parabola degenerate. Therefore, the equation given does not directly correspond to a valid parabola with the point (-4a, 0).
Determining the Tangent and Normal Equations
Despite the issue with the point, let's assume the problem intends to find the tangent and normal to the parabola x^2 16a at a general point on the x-axis, say (4a, 0).
Equation of the Tangent at (4a, 0)
The equation of the tangent to a parabola x^2 4ay at a point (x_1, y_1) is given by:
Equation of the tangent: x x_1 2a(y y_1)
For the point (4a, 0) on the parabola:
Tangent: 4a x 2a(y 0)
4a x 2ay
2x y
Thus, the equation of the tangent at the point (4a, 0) is:
y 2x
Equation of the Normal at (4a, 0)
The slope of the tangent at a point on a parabola is given by the derivative of the function. For the parabola x^2 16a, the derivative is:
dy/dx x / (4a)
At the point (4a, 0), the slope of the tangent is:
dy/dx 4a / (4a) 1
The slope of the normal is the negative reciprocal of the slope of the tangent, which is:
-1 / 1 -1
The equation of the normal at the point (4a, 0) with slope -1 is:
y - 0 -1(x - 4a)
y -x 4a
Thus, the equation of the normal at the point (4a, 0) is:
y -x 4a
Conclusion
In summary, the point with coordinate -4a does not lie on the parabola given by x^2 16a. However, based on the typical parabolic equation and the standard form, we have derived the equations for the tangent and normal at the point (4a, 0) on the parabola x^2 16a.
Key Takeaways:
- The tangent at (4a, 0) is y 2x.
- The normal at (4a, 0) is y -x 4a.