Conic Sections

Let tanα, tanβ and tanγ; α,β,γ ≠ (2n-1)π/2, n ∈ N be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of ΔABC coincides with origin and its orthocenter lies on y-axis, then the value of (cos(3α) + cos(3β) + cos(3γ)) / (cosα cosβ cosγ) is equal to ______.

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Contributor-Level 10

If the orthocenter and circumcenter of a triangle both lie on the y-axis, the centroid also lies on the y-axis.
The x-coordinate of the centroid is (x1 + x2 + x3) / 3. If the vertices are (cos α, sin α), (cos β, sin β), (cos γ, sin γ), then the x-coordinate of the centroid is (cos α + cos β + cos γ) / 3.
Since the centroid lies on the y-axis, its x-coordinate is 0.
cos α + cos β + cos γ = 0

Using the identity: If a + b + c = 0, then a^3 + b^3 + c^3 = 3abc.
Let a = cos α, b = cos β, c = cos γ.
Then, cos^3 α + cos^3 β + cos^3 γ = 3 * cos α * cos β * cos γ.

We need to evaluate the expression:
(cos 3α + cos 3β + cos 3γ) / (

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New answer posted

a month ago

Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be 1/2 and probability of occurrence of 0 at the odd place be 1/3. Then the probability that '10' is followed by 01 is equal to:

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Contributor-Level 10

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.

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New answer posted

a month ago

Let O be the origin. Let OP = xi + yj - k and OQ = -i + 2j + 3xk, x, y ∈ R, x > 0, be such that |PQ| = √20 and the vector OP is perpendicular to OQ. If OR = 3i + zj - 7k, z ∈ R, is coplanar with OP and OQ, then the value of x² + y² + z² is equal to:

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Contributor-Level 10

Given vectors OP = xi + yj - k and OQ = -i + 2j + 3xk.
PQ = OQ - OP = (-1 - x)i + (2 - y)j + (3x + 1)k
Given |PQ| = √20, so |PQ|² = 20.
(-1 - x)² + (2 - y)² + (3x + 1)² = 20
(1 + x)² + (2 - y)² + (3x + 1)² = 20 .(i)
- Given OP ⊥ OQ, so OP · OQ = 0.
  (x)(-1) + (y)(2) + (-1)(3x) = 0
  -x + 2y - 3x = 0 ⇒ -4x + 2y = 0 ⇒ y = 2x .(ii)
Substitute (ii) into (i):
(1 + x)² + (2 - 2x)² + (3x + 1)² = 20
1 + 2x + x² + 4 - 8x + 4x² + 9x² + 6x + 1 = 20
14x² = 14 ⇒ x² = 1 ⇒ x = ±1.
When x = 1, y = 2. When x = -1, y = -2.
So, (x, y) can be (1, 2) or (-1, -2).
- Given that OR, OP, and OQ are coplanar, their scalar triple product is 0: [OR OP OQ]

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New answer posted

a month ago

If lim (x→0) [sin⁻¹x - tan⁻¹x] / 3x³ is equal to L, then the value of (6L + 1) is:

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Contributor-Level 10

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.

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New answer posted

a month ago

The differential equation satisfied by the system of parabolas y² = 4a(x + a) is:

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Contributor-Level 10

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

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New question posted

a month ago

For the four circles M, N, O and P, following four equations are given:
Circle M: x² + y² = 1
Circle N: x² + y² - 2x = 0
Circle O: x² + y² - 2x - 2y + 1 = 0
Circle P: x² + y² - 2y = 0
If the centre of circle M is joined with centre of the circle N, further centre of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a:

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New answer posted

a month ago

The minimum distance between any two points P₁ and P₂ while considering point P₁ on one circle and point P₂ on the other circle for the given circles' equations x² + y² - 10x - 10y + 41 = 0, x² + y² - 24x - 10y + 160 = 0 is.

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Contributor-Level 10

Kindly consider the following figure

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New answer posted

a month ago

The line 2x - y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x - 2y = 4. Then, the radius of the circle is:

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Contributor-Level 10

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.

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New answer posted

a month ago

If the co-ordinates of two points A and B are (√7, 0) and (−√7, 0) respectively and P is any point on the conic 9x² + 16y² = 144, then PA + PB is equal to:

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Contributor-Level 10

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

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New answer posted

a month ago

If the common tangent to the parabolas y² = 4x and x² = 4y also touches the circle, x² + y² = c², then c is equal to

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