Ncert Solutions Maths class 11th

The equation of the tangent to the ellipse x²/27 + y² = 1 at the point (3√3 cosθ, sinθ) is:
x (3√3 cosθ)/27 + y (sinθ)/1 = 1 ⇒ x/ (3√3) cosθ + y sinθ = 1.
To find the intercepts on the axes:
x-intercept (set y=0): x = 3√3 / cosθ = 3√3 secθ.
y-intercept (set x=0): y = 1 / sinθ = cosecθ.
The sum of the intercepts is z (θ) = 3√3 secθ + cosecθ.
To find the minimum value of z, we differentiate with respect to θ and set it to zero:
dz/dθ = 3√3 secθ tanθ - cosecθ cotθ = 0.
3√3 (sinθ/cos²θ) = cosθ/sin²θ.
3√3 sin³θ = cos³θ ⇒ tan³θ = 1/ (3√3).
⇒ tanθ = 1/√3.
Since θ ∈ (0, π/2), the solution is θ = π/6.
A check of the second derivative confirms this is a minimum. Thus, z is minimum for θ = π/6.

Given the equation 15 sin? + 10 cos? = 6.
Divide by cos? : 15 tan? + 10 = 6 sec?
Using sec²? = 1 + tan²? , we get sec? = (1 + tan²? )² = 1 + 2tan²? + tan?
15 tan? + 10 = 6 (1 + 2tan²? + tan? ).
15 tan? + 10 = 6 + 12tan²? + 6tan?
9 tan? - 12 tan²? + 4 = 0.
This is a quadratic in tan²? : (3 tan²? - 2)² = 0.
? 3 tan²? = 2? tan²? = 2/3.
From this, we find sin²? and cos²? If tan²? = 2/3, then sin²? = 2/5 and cos²? = 3/5.
Also, sec²? = 1 + tan²? = 5/3 and cosec²? = 1 + cot²? = 1 + 3/2 = 5/2.
The expression to evaluate is 27 sec? + 8 cosec? = 27 (sec²? )³ + 8 (cosec²? )³.
= 27 (5/3)³ + 8 (5/2)³ = 27 (125/27) + 8 (125/8) = 125 + 125 = 250.

The equation is for a hyperbola: x²/4 - y²/2 = 1.
The eccentricity e is given by e = √ (1 + b²/a²) = √ (1 + 2/4) = √6/2.
The focus F is at (ae, 0), which is (2 * √6/2, 0) = (√6, 0).
The equation of the tangent at a point P (x? , y? ) is xx? /a² - yy? /b² = 1.
The equation of the tangent at P is given as x - (√6 y)/2 = 1.
This tangent cuts the x-axis (y=0) at x=1, so Q is (1, 0).
The latus rectum is the line x = ae = √6.
To find the point R where the tangent intersects the latus rectum, we substitute x=√6 into the tangent equation:
√6 - (√6 y)/2 = 1 ⇒ √6 - 1 = (√6 y)/2 ⇒ y = 2 (√6 - 1)/√6.
The vertices of the triangle QFR are Q (1,0), F (√6,0), and R (√6, 2 (√6 - 1)/√6).
The base QF has length √6 - 1.
The height is the y-coordinate of R.
Area of ΔQFR = 1/2 * base * height = 1/2 * (√6 - 1) * [2 (√6 - 1)/√6].
Area = (√6 - 1)² / √6 = (6 - 2√6 + 1) / √6 = (7 - 2√6) / √6.

Let x = m(a + λb).
Given m(a + λb) ⋅ (3i + 2j - k) = 0, which leads to λ = -3/8.

The projection of vector x on vector a is given by x ⋅ â, where â is the unit vector of a.
Projection = (x ⋅ a) / |a| = 17√6 / 2
x ⋅ a = (m(a + λb)) ⋅ a = m(a ⋅ a + λ(b ⋅ a)) = m(|a|^2 + λ(b ⋅ a))

The provided text simplifies this to:
m(6 - 3/8 * (-1)) = 17√6 / 2
m * (51/8) = 17 * 6 / 2 (The text seems to have a typo 17x6/2 instead of 17√6 / 2)
Assuming it is 17 * 6 / 2, m * 51/8 = 51, so m = 8.

x = 8(a + (-3/8)b) = 8a - 3b
x = 8( (13/8)i - (14/8)j + (11/8)k ) (The vectors a and b are not fully defined in the provided text)
The final vector is given as x = 13i - 14j + 11k.

|x|^2 = 13^2 + (-14)^2 + 11^2 = 169 + 196 + 121 = 486.

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) / (cos α * cos β * cos γ)

Using the identity cos 3θ = 4cos^3 θ - 3cos θ:
cos 3α + cos 3β + cos 3γ = (4cos^3 α - 3cos α) + (4cos^3 β - 3cos β) + (4cos^3 γ - 3cos γ)
= 4(cos^3 α + cos^3 β + cos^3 γ) - 3(cos α + cos β + cos γ)

Substitute the known values:
= 4(3 * cos α * cos β * cos γ) - 3(0)
= 12 * cos α * cos β * cos γ

So, the expression becomes:
(12 * cos α * cos β * cos γ) / (cos α * cos β * cos γ) = 12.
The value of the expression squared is 12^2 = 144.