### Math Notes

Subjects

#### Analytic Geometry Solutions

##### Topics || Problems

Find the equation of the parabola with axis parallel to 0x and passing through (5, 4) (11, -2),(21, -4,).

Since the parabola is parallel to 0x, the curve must be opening to either left or right.

To find the equation using the given points;

• Create equations - $$(y-k)^2 = 4a(x-h)$$
• Find the values of a, h and k

Create Equations

• $$(4-k)^2 =4a(5-h)$$, points$$(5,4)$$
• $$(-2-k)^2 =4a(11-h)$$, points$$(11,-2)$$
• $$(-4-k)^2 =4a(21-h)$$, points$$(21,-4)$$

Use the equations to calculate $$a, h ~and ~ k$$

$$(4-k)^2 =4a(5-h)$$

$$16 - 8k + k^2 = 20a -4ah$$ $$eq. ~1$$

$$(-2-k)^2 =4a(11-h)$$

$$4 +4k + k^2 = 44a -4ah$$ $$eq. ~2$$

$$(-4-k)^2 =4a(21-h)$$

$$16+8k + k^2 = 84a -4ah$$ $$eq. ~3$$

Remove $$h$$ using equations $$1$$ and $$2$$

\begin{align}(16 - 8k + k^2)-(4 +4k + k^2) \\= (20a -4ah)-(44a -4ah)\end{align}

$$12-12k = -24a$$ $$eq.~4$$

Remove $$h$$ using equations $$2$$ and $$3$$

\begin{align}(4 +4k + k^2) - (16+8k + k^2 ) \\= (44a -4ah) - (84a -4ah)\end{align}

$$-12 - 4k = -40a$$ $$eq.~5$$

Find $$k ~ and ~ a$$ using equations $$5~ and ~4$$

$$12-12k = -24a)$$ $$eq.~4$$

$$36 + 12k = 120a~~$$ $$eq.~5$$ multiplied by $$-3$$

$$48 = 96a$$

$$a = \frac{1}{2}$$

Solve $$k$$

$$12-12k = -24 \frac{1}{2}$$

$$k=2$$

Use any equations $$(1-3)$$ to solve $$h$$.

$$(4-2)^2 = 4(0.5)(5-h)$$

$$h = 3$$

The equation:

$$(y-2)^2 = 2(x-3)$$ Answer

It can also be written in general form.

$$y^2-4y + 4 = 2x-6$$

$$y^2-4y-2x+10=0$$ Answer