The line with the equation \(x - 3y - 27 = 0\) meets the parabola \(y^2 = 4x\) at two points. Find their coordinates.

x − 3 y − 27 = 0 y 2 = 4 x x − 3 y − 27 = 0 x = 4 y 2 ​ 4 y 2 ​ − 3 y − 27 = 0 y 2 − 12 y − 108 = 0 y 2 − 12 y + 36 − 144 = 0 ( y − 6 ) 2 = 144 y − 6 = − 12 ∨ y − 6 = 12 y = − 6 ∨ y = 18 x = 4 ( − 6 ) 2 ​ ∨ x = 4 1 8 2 ​ x = 4 36 ​ ∨ x = 4 324 ​ x = 9 ∨ x = 81 ( 9 , − 6 ) , ( 81 , 18 ) ​

Answered by konrad509 | 2024-06-10

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 = √(4 9) = √36 = ±6 When x = 81, y = √(4 81) = √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)

Answered by DavidOrrell | 2024-06-10

The points where the line x − 3 y − 27 = 0 intersects the parabola y 2 = 4 x are ( 9 , − 6 ) and ( 81 , 18 ) . We find these intersections by substituting the line's equation into the parabola's equation to derive a quadratic equation in y , which we then solve. Plugging the corresponding y values back into the line's equation gives us the complete coordinates of the intersection points.
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Answered by konrad509 | 2024-12-26