Conic Sections

P(O at even place) = 1/2, P('O' at odd place) = 1/3

P(1 at even place) = 1 - 1/2 = 1/2

P(1 at odd place) = 1 - 1/3 = 2/3

The probability that 10 is followed by 01 is given by a sequence of probabilities, which seems to represent a specific arrangement or transition. The calculation is:
P(10 is followed by 01) = (2/3) * (1/2) * (1/3) * (1/2) * (1/2) * (1/3) * (1/2) * (2/3) = 1/18 + 1/18 = 1/9.
The calculation seems incomplete or context is missing.

Evaluate the limit:
L = lim (x→0) [sin? ¹ (x) - tan? ¹ (x)] / 3x³

Using Taylor series expansions around x=0:
sin? ¹ (x) = x + x³/6 + O (x? )
tan? ¹ (x) = x - x³/3 + O (x? )

L = lim (x→0) [ (x + x³/6) - (x - x³/3) ] / 3x³
L = lim (x→0) [ x³/6 + x³/3 ] / 3x³
L = lim (x→0) [ (1/6 + 1/3)x³ ] / 3x³
L = (1/2) / 3 = 1/6

The solution shows 3L = 1/2, which is correct. And 6L = 1, also correct.
The final line 6L+1=2 implies 6L=1, confirming the result.

Given the family of parabolas y² = 4a (x+a).
Differentiate with respect to x:
2y (dy/dx) = 4a
a = (y/2) (dy/dx)

Substitute a back into the original equation:
y² = 4 * (y/2) (dy/dx) * [x + (y/2) (dy/dx)]
y² = 2y (dy/dx) * [x + (y/2) (dy/dx)]
y = 2 (dy/dx) * [x + (y/2) (dy/dx)]
y = 2x (dy/dx) + y (dy/dx)²
y (dy/dx)² + 2x (dy/dx) - y = 0

Kindly consider the following figure

Given equation of tangent is 2x - y + 1 = 0
equation of normal is x + 2y = 12
Solving with x - 2y = 4 we get centre at (6,2) radius = √ (36 + 9) = √45 = 3√5.

For ellipse x²/16 + y²/9 = 1, a=4, b=3, e = √ (1 - 9/16) = √7/4
A and B are foci then PA + PB = 2a = 2 (4) = 8

y² = 4x and x² = 4y
any tangent of y² = 4x is y = mx + 1/m
it also tangent for x² = 4y
1/m = -m² ⇒ m = -1
∴ common tangent is y = -x - 1, it also touches x² + y² = c²
∴ 1 = c² ⋅ (1+1) ⇒ c² = 1/2

For the parabola y = x², the tangent at (2,4) is given by (y+4)/2 = 2x, which simplifies to 4x - y - 4 = 0.
The equation of a circle touching the line 4x - y - 4 = 0 at the point (2,4) is
(x-2)² + (y-4)² + λ (4x-y-4) = 0.
It passes through (0,1).
∴ (0-2)² + (1-4)² + λ (4 (0) - 1 - 4) = 0
4 + 9 + λ (-5) = 0 ⇒ 13 = 5λ ⇒ λ = 13/5
∴ the circle is x² - 4x + 4 + y² - 8y + 16 + (13/5) (4x-y-4) = 0
x² + y² + (-4 + 52/5)x + (-8 - 13/5)y + (20 - 52/5) = 0
x² + y² + (32/5)x - (53/5)y + 48/5 = 0
∴ centre is (-g, -f) = (-16/5, 53/10).

Equation of normal to the ellipse x²/a² + y²/b² = 1 at (x? , y? ) is a²x/x? - b²y/y? = a² - b².
At the point (ae, b²/a):
a²x/ (ae) - b²y/ (b²/a) = a² - b²
It passes through (0, -b).
a² (0)/ (ae) - b² (-b)/ (b²/a) = a² - b²
ab = a² - b²
Since b² = a² (1-e²), a²-b² = a²e².
ab = a²e²
a²b² = a? e?
a² (a² (1-e²) = a? e?
1 - e² = e?
e? + e² - 1 = 0