1. ## Coordinates Question

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

Attempt:

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

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

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

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

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

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

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

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

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

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

Where did I go wrong?

2. Hi

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

Attempt:

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

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

^^

3. For your problem, you have to solve simultaneously these equations :

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

Square the second equation

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

Substitute the first equation into this one :

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

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

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

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