Question

Is the point (11,5) on the parabola? Show or explain your work.

Answer 1

Answer:

The point (11, 5) is NOT on the parabola.

Step-by-step explanation:

Let $(x_0,y_0)$ be any point on the parabola

Let $(a, b)$ be the focus

Let $y=c$ be the the directrix

Distance between $(x_0,y_0)$ and the focus: $\sqrt{(x_0-a)^2+(y_0-b)^2}$

Distance between $(x_0,y_0)$ and the directrix is $|y_0-c|$

Given:

• focus  (6, 4)
• directrix  y = 0

Therefore:

Distance between $(x_0,y_0)$ and the focus: $\sqrt{(x_0-6)^2+(y_0-4)^2}$

Distance between $(x_0,y_0)$ and the directrix is $|y_0-0|$

Equate the two expressions and solve for $y_0$:

     $\sqrt{(x_0-6)^2+(y_0-4)^2}=|y_0-0|$

$\implies (x_0-6)^2+(y_0-4)^}=(y_0-0)^2$

$\implies {x_0}^2-12x_0+36+{y_0}^2-8y_0+16={y_0}^2$

$\implies {x_0}^2-12x_0-8y_0+52=0$

$\implies 8y_0={x_0}^2-12x_0+52$

$\implies y_0=\dfrac18{x_0}^2-\dfrac{12}{8}x_0+\dfrac{52}{8}$

$\implies y_0=\dfrac18{x_0}^2-\dfrac32x_0+\dfrac{13}{2}$

Now rewrite with (x, y):

$\implies y=\dfrac18x^2-\dfrac32x+\dfrac{13}{2}$

Therefore, this is the equation of the parabola

To determine if the point (11, 5) is on the parabola, input x = 11 into the equation:

$\implies y=\dfrac18(11)^2-\dfrac32(11)+\dfrac{13}{2}$

$\implies y=\dfrac{41}{8}$

So the point (11, 5) is NOT on the parabola.