Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

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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alok kumar singh | Contributor-Level 10

a month ago

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 &alph

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Or 4 (1 – e²) = e²

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$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

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Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

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$\frac{a^{2}}{b^{2}} = \frac{1}{2}$

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$\frac{144}{25} + \frac{256}{25} \times \frac{1}{2} = a^{2}$

$\frac{(128 + 144)}{25} = a^{2} \Rightarrow \frac{272}{25} = a^{2}$

-> $b^{2} = \frac{2 \times 272}{25}$

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(Latus rectum)²

$= \frac{4 b^{4}}{a^{2}} = 4 (\frac{b^{2}}{a^{2}}) b^{2} = \frac{8 \times 272 \times 2}{25} = \frac{4352}{25}$

less

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Vishal Baghel

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48 = 16 (e2 – 1)

$e = \sqrt[]{4} = 2$

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