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ollegr [7]
3 years ago
14

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

Mathematics
2 answers:
lesantik [10]3 years ago
4 0
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)}

sashaice [31]3 years ago
4 0
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 = <span>√4x
When x = 9, y = </span>√(4*9) = √36 = <span>±6
When x = 81, y = </span>√(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 <span>≠ 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 </span><span>≠</span> 0

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





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