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secondary 3 | A Maths
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Date Posted: 4 years ago
Views: 311
y = 0.5x + p
x² + y² = 8p
*keyword: touches
x² + (0.5x + p)² = 8p
x² + 0.25x² + px + p² = 8p
5x²/4 + px + (p²-8p) = 0
*"touches" means one root
=> p² - 4 (5/4) (p²-8p) = 0
=> p² - 5p² + 40p = 0
=> p² - 10p = 0
=> p(p - 10) = 0
=> p = 0 or p = 10
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.
.
x² + y² = 8p
*keyword: touches
x² + (0.5x + p)² = 8p
x² + 0.25x² + px + p² = 8p
5x²/4 + px + (p²-8p) = 0
*"touches" means one root
=> p² - 4 (5/4) (p²-8p) = 0
=> p² - 5p² + 40p = 0
=> p² - 10p = 0
=> p(p - 10) = 0
=> p = 0 or p = 10
.
.
.
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I have some reservations on the solution p = 0, as the case p = 0 leads to the curve equation being x2 + y2 = 0 which is not exactly a curve at all.
We know that perfect squares have a minimum value of 0, so the only way that x2 + y2 can equal 0 is when both x and y are 0, and thus the “curve” is actually only one single coordinate and not exactly a curve.
(For your information, x2 + y2 = 0 represents a “circle” centred at (0, 0) having a radius of “0” units; you will learn circles late in Sec 3 A Maths)
We know that perfect squares have a minimum value of 0, so the only way that x2 + y2 can equal 0 is when both x and y are 0, and thus the “curve” is actually only one single coordinate and not exactly a curve.
(For your information, x2 + y2 = 0 represents a “circle” centred at (0, 0) having a radius of “0” units; you will learn circles late in Sec 3 A Maths)
Date Posted:
4 years ago
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