# Coordinates Question

• May 21st 2008, 12:46 AM
looi76
Coordinates Question
Question:
The curve $\displaystyle y^2 = 12x$ intersects the line $\displaystyle 3y = 4x + 6$ at two points. Find the distance between the two points.

Attempt:

$\displaystyle y^2 = 12x$
$\displaystyle y = (12x)^{\frac{1}{2}}$

$\displaystyle 3y = 4x + 6$
$\displaystyle y = \frac{4}{3}x + 2$

$\displaystyle (12x)^{\frac{1}{2}} = \frac{4}{3}x + 2$
$\displaystyle 12x = \left(\frac{4}{3}x + 2\right)^2$

$\displaystyle 12x = \left(\frac{4}{3}x + 2\right)\left(\frac{4}{3}x + 2\right)$

$\displaystyle 12x = \frac{16}{9}^2 + \frac{16}{3}x + 4$

$\displaystyle = \frac{16}{9}^2 + \frac{16}{3}x - 12x + 4$

$\displaystyle = \frac{16}{9}^2 - \frac{20}{3}x + 4$

$\displaystyle a = \frac{16}{9} \ , \ b = -\frac{20}{3} \ , \ c = 4$

$\displaystyle x = \frac{-b \pm \sqrt(b^2 - 4ac)}{2a}$

$\displaystyle x = \frac{-(-\frac{20}{3}) \pm \sqrt((-\frac{20}{3})^2 - 4 \times \frac{16}{9} \times 4)}{2 \times \frac{16}{9}}$

Where did I go wrong?(Doh)
• May 21st 2008, 12:50 AM
Moo
Hi :)

Quote:

Originally Posted by looi76
Question:
The curve $\displaystyle y^2 = 12x$ intersects the line $\displaystyle 3y = 4x + 6$ at two points. Find the distance between the two points.

Attempt:

$\displaystyle y^2 = 12x$
$\displaystyle y = (12x)^{\frac{1}{2}}$

[...]

Remember that if $\displaystyle a^2=b^2$, then $\displaystyle a={\color{red}\pm} b$

^^
• May 21st 2008, 12:55 AM
Moo
For your problem, you have to solve simultaneously these equations :

$\displaystyle y^2=12x$
$\displaystyle 3y=4x+6$

Square the second equation (Sun)

$\displaystyle 9y^2=16x^2+48x+36$

Substitute the first equation into this one :

$\displaystyle 9(12x)=16x^2+48x+36$

$\displaystyle 16x^2-60x+36=0 \longleftrightarrow 4x^2-15x+9=0$

$\displaystyle 4x^2-15x+9=(x-3)(4x-3)=0$

Therefore, $\displaystyle x=3$ or $\displaystyle x=\frac 34$

:)