Parabola Problems and Solutions

In mathematics, the parabola is a conic section, the intersection of a right circular conical surface and a plane to a generating straight line of that surface. Given a point (the focus) and a corresponding line (the directrix) on the plane, the locus of points in that plane that are equidistant from them is a parabola (Source: Wikipedia). In this topic we discuss about parabola problems and solutions, parabola example problems

Parabola Grapher
Parabola Example Problems with Solutions:

Problem: 1

What is the minimum value of the expression 2x2 – 20x + 17?

Solution:Parabola Example 1 Diagram

Consider the function y = 2x2 – 20x + 17. This function is defined by a second degree equation. This xo-efficient of its x2 term is positive. Hence the curve is a parabola opening up ward.

`(-coefficient fo X term)/(2.coefficient fo X^2 term)` = `-b/(2a)` =` (-(-20))/(2(2))` = `20/4` = 5.

For x = 5, y = 2(5)2 – 20(5) + 17 = - 33. Therefore the minimum value of the expression 2x2- 20x + 17 for any value of x is – 33. This minimum value is assumed only when x = 5.
Problem: 2

Find the coordinates of maximum point of the curve y = - 3x2 – 12x + 5, and locate the axis of symmetry.

Solution:

The curve is defined by a second degree equation. The coefficient of x2 term is negative.

`(-coefficient fo X term)/(2.coefficient fo X^2 term)` = `-b/(2a)` = `(-(-12))/(2(-3))` = `12/-6` = -2.

For x = -2, y = -3 (-2)2 – 12(- 2) + 5 = 17. Hence the coordinates of the vertex are (- 2, 17). The curve is symmetric with respect to the vertical line through its vertex, through the point (-2, 17), i.e., the line x = -2


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Problem: 3

If a parabolic reflector is 16 cm in diameter and 4cm deep, find the focus.

Solution:

let POQ be the vertical section of the reflector. Mid - point of PQ is M. Let OX be along OM and OY parallel to MP.

Let the equation of the parabola be y2 = 4ax.

The coordinates of P are (4, 8)

(8)2= 4a (4) or a = 4

Focus = (a, 0) = (4, 0).

Focus coincides with M, the mid-point of PQ