coordinate geometry

[dull current]

how to prove

[spare saddle]

draw it

[dull current]

and then what

[undone quiver]

what is the answer

[dull current]

i don’t know

[dull current]

???

[stray mantle]

We can parametrise the quarter circle as follows:
x = cos(t)
y = sin(t)
0 ≤ t ≤ π/2
Then:
dx/dt = -sin(t)
dy/dt = cos(t)
dy/dx = (dy/dt)/(dx/dt) = -cot(t)
So, we have a line passsing through (cos(t), sin(t)) with slope -cot(t). We can find its equation.
y = -cot(t)(x - cos(t)) + sin(t)
So, now find the area of the triangle bounded by this line and the coordinate axes and set it to be equal to, say, S. Then find the value of t for which it is satisfied. Finally, substitute S = 2.

[dull current]

thanks!

[signal estuary]

@dull current has given 1 rep to @stray mantle

[dull current]

what did you define t as tho?

[stray mantle]

A parameter for out curve - a quarter-circle.

[dull current]

stilll slightly confused abouot last part

how do i get the gradient from this

[spare saddle]

y = mx + c
y-intercept = c (set x = 0)
x-intercept = -c/m (set y = 0)
Area = (1/2)(c)(-c/m) = 2
-c²/2m = 2
-c² = 4m
4m + c² = 0 (1)
x² + y² = 1
x² + (mx + c)² = 1
x² + m²x² + 2cmx + c² = 1
(1 + m²)x² + (2cm)x + (c² - 1) = 0
tangent -> D = 0
(2cm)² - 4(1 + m²)(c² - 1) = 0
4c²m² - 4(c² - 1 + c²m² - m²) = 0
4c²m² - 4c² + 4 - 4c²m² + 4m² = 0
4m² - 4c² = -4
m² - c² = -1 (2)
(1) + (2)
4m + m² = -1
m² + 4m + 1 = 0
m = (-4 ± sqrt(4² - 4(1)(1))/2(1)
= (-4 ± sqrt(12))/2
= (-4 ± 2sqrt(3))/2
= -2 ± sqrt(3)
m1 = -2 + sqrt(3)
m2 = -2 - sqrt(3)

[stray mantle]

If you have a parametric curve x = x(t),y = y(t), then dy/dx = y'(t)/x'(t).

[dull current]

ok got it

thanks guys

[stray mantle]

You're welcome!