A square ABCD has all its vertices on the curve x2y2 = 1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is ______.

A square ABCD has all its vertices on the curve x²y² = 1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is ______.

6 Views|Posted 5 months ago

Asked by Shiksha User

No answers yet.

Similar Questions for you

Let c be a vector perpendicular to the vectors a=i+j-k and b=i+2j+k. If c. (i + j + 3k) = 8 then the value of c. (a x b) is equal to......... c. (a x b) is equal to.........

Alok Kumar Singh

c = λ (a x b).
a = I + j - k
b = I + 2j + k
a x b = | I j k |
| 1 -1 |
| 1 2 1 |
= I (1 - (-2) - j (1 - (-1) + k (2-1) = 3i - 2j + k.
c = λ (3i - 2j + k).
Given c ⋅ (i + j + 3k) = 8.
λ (3i - 2j + k) ⋅ (i + j + 3k) = 8
λ (3 - 2 + 3) = 8 => 4λ = 8 => λ = 2.
c = 2 (a x b).
We need to find c ⋅ (a x b).
c ⋅ (a x b) =

Vishal Baghel

Kindly consider the following figure

Let x₀ be the point of local maxima of f(x) = a · (b × c) where a = xî − 2ĵ + 3k, b = −2î + xĵ − k and c = 7î – 2ĵ + xk. Then the value of a · b + b · c + c · a at x = x₀ is

Alok Kumar Singh

f (x) = a? ⋅ (b? × c? ) = |x -2 3; -2 x -1; 7 -2 x|
= x³ - 27x + 26
f' (x) = 3x² - 27 = 0 ⇒ x = ±3 and f' (-3) < 0
⇒ local maxima at x = x? = -3
Thus, a? = -3i? - 2j? + 3k? , b? = 2i? - 3j? - k? , and c? = 7i? - 2j? - 3k?
⇒ a? ⋅ b? + b? ⋅ c? + c? ⋅ a? = 9 - 5 - 26 = -22