the line with equation x-3y-27=0 meets the parabola y²=4x at two points. Find their coordinates

\(x-3y-27=0\\ y^2=4x\\\\ x-3y-27=0\\ x=\frac{y^2}{4}\\\\ \frac{y^2}{4}-3y-27=0\\ y^2-12y-108=0\\ y^2-12y+36-144=0\\ (y-6)^2=144\\ y-6=-12 \vee y-6=12\\ y=-6 \vee y=18\\\\ x=\frac{(-6)^2}{4} \vee x=\frac{18^2}{4}\\ x=\frac{36}{4} \vee x=\frac{324}{4}\\ x=9 \vee x=81\\\\ \boxed{(9,-6),(81,18)}\)

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DavidOrrell DavidOrrell · Answer 2 · 2014-10-24T05:56:55-04:00

First, rearrange the first equation:

3y = x - 27
y = x/3 - 9

Square this equation to make y^2 the subject:

y^2 = (x/3 - 9)^2 = (x/3 - 9)(x/3 - 9) = (x^2)/9 - 3x - 3x + 81 = (x^2)/9 - 6x + 81

Now you can substitute this for y^2 in the second equation, then rearrange into the form ax^2 + bx + c = 0:

(x^2)/9 - 6x + 81 = 4x
(x^2)/9 - 10x + 81 = 0
x^2 - 90x + 729 = 0

Factorise the equation, then equate to zero and zolve:

(x - 9)(x - 81) = 0

x - 9 = 0 --> x = 9
x - 81 = 0 --> x = 81

Using these x values, find the corresponding y values:

y^2 = 4x ∴ y = √4x
When x = 9, y = √(49) = √36 = ±6
When x = 81, y = √(481) = √324 = ±18

Now we need to test whether each y co-ordinate is positive or negative:

When x = 9 and y = 6: x - 3y - 27 = 9 - 18 - 27 ≠ 0
When x = 9 and y = -6: x - 3y - 27 = 9 + 18 - 27 = 0

When x = 81 and y = 18: x - 3y - 27 = 81 - 54 - 27 = 0
When x = 81 and y = -18: x - 3y - 27 = 81 + 54 - 27 ≠ 0

Therefore, the co-ordinates of the points of intersection are (9, -6) and (81, 18)