Chapter 3.4, Problem 25HP

To determine

To find: The values of a, b, and c and the general form of the equation.

Answer to Problem 25HP

  $a=313,b=-504,c=-144$

The general form of the equation of the parabola is $y^2=313x^2-504x-144$.

Explanation of Solution

Given information: The general form of an equation for a parabola is $y^2=a{x}^2+bx+c$, where $(x,y)$ is a point on the parabola. Three points on a parabola are $(2,-10),(-5,-101)$ and $(6,-90)$.

Calculation: The general form of an equation for a parabola is $y^2=a{x}^2+bx+c$

Putting $(2,-10)$ on the equation,

  ${(-10)}^2=a{(2)}^2+b(2)+c$

  $\Rightarrow 4a+2b+c=\mathrm{100..............................}(1)$

Putting $(-5,-101)$ on the equation,

  ${(-101)}^2=a{(-5)}^2+b(-5)+c$

  $\Rightarrow 25a-5b+c=\mathrm{10201..............................}(2)$

Putting $(6,-90)$ on the equation,

  ${(-90)}^2=a{(6)}^2+b(6)+c$

  $\Rightarrow 36a+6b+c=\mathrm{8100..............................}(3)$

Subtracting equation (1) from equation (2),

  $(25a-5b+c)-(4a+2b+c)=10201-100$

  $\Rightarrow 25a-5b+c-4a-2b-c=10101$

  $\Rightarrow 21a-7b=10101$

  $\Rightarrow 3a-b=\mathrm{1443..............................}(4)$

Subtracting equation (1) from equation (3),

  $(36a+6b+c)-(4a+2b+c)=8100-100$

  $\Rightarrow 36a+6b+c-4a-2b-c=8000$

  $\Rightarrow 32a+4b=8000$

  $\Rightarrow 8a+b=\mathrm{2000..............................}(5)$

Adding equation (4) and (5),

  $(3a-b)+(8a+b)=1443+2000$

  $\Rightarrow 3a-b+8a+b=3443$

  $\Rightarrow 11a=3443$

  $\Rightarrow a=313$

Putting $a=313$ in equation(4),

  $3a-b=1443$

  $\Rightarrow 3(313)-b=1443$

  $\Rightarrow b=939-1443$

  $\Rightarrow b=-504$

Putting $a=313,b=-504$ in equation (1),

  $4a+2b+c=100$

  $\Rightarrow 4(313)+2(-504)+c=100$

  $\Rightarrow 1252-1008+c=100$

  $\Rightarrow 244+c=100$

  $\Rightarrow c=100-244$

  $\Rightarrow c=-144$

Thus, $a=313,b=-504,c=-144$

The general form of the equation of the parabola is,

  $y^2=313x^2-504x-144$