The minimum value of the function f(x,y) = x² + y² on the hyperbola xy = 5 is 10. The detailed solution for the same is added below:

Minimize f (x, y) = x² + y² on the hyperbola xy = 5.

Solution:

f(X, Y) = X² + Y² ⇢ (i)

XY = 5

Y = 5/X ⇢ (ii)

f(X) = X²+ (5/X)² ⇢ (by putting equation1 into equation 2)

f'(X) = 2X + 25(-2/X³) ⇢ (1st derivative of equation) 

Now, 2X + 25(-2/X³) = 0

2X – 50/X³ = 0

2X⁴ – 50 = 0

X⁴ = 50/2

X⁴ = 25

X² = ±√25

X² = ±5 ⇢ [as X² can’t be -ve]

X = ±√5  

Y = 5/ ±√5 ⇢ [put the value of X into equation (ii)

Y =  ±(√5 × √5)/ √5

Y = ±√5

(X, Y) = (√5 ,√5) & (-√5, -√5)

Minimum value: X² + Y² = (±√5 )²+ (±√5)² = 10

Similar Problems

Problem 1: Find the position of the point (6, -5) relative to the hyperbola x²/a² – y²/b² = 1?

Solution: 

We know that point p(x₁, y₂) lies 

Outside, x²/a² – y²/b² – 1 < 0

Inside, x²/a² – y²/b² – 1 < 0

So, x₁²/a² – y²₁/b2 – 1 

= 6²/9 – (-5)²/25 – 1

= 36/9 – 25/25 – 1

 = 2

Since 2 > 0

Therefore point (6, -5) lies inside the  given equation of hyperbola.

Problem 2: Find the eccentricity of the hyperbola 4x² – 9y² -8x = 32?

Solution: 

4x² – 9y² – 8x = 32

4(x² – 2x ) – 9y² = 32

4(x² – 2x + 1) – 9y² = 32 + 4  [adding 4 both sides]

4(x² – 2x + 1) – 9y² = 36

[(x -1)²]/9 – [y²]/4 = 1 

a² = 9 , b² = 4

e = [1 + b²/a²]^1/2

e = (1 + 4/9)^1/2

e = (13/9)^1/2

e = (√13)/3

Therefore, e = (√13)/3

Problem 3: If the center, vertex, and locus of a hyperbola be (0, 0),(4, 0), and (6, 0) respectively, then find the equation of the hyperbola?

Solution: 

Center (0, 0)

Vertex (4, 0)

Focus (6, 0)

So, a = 4

ae = 6

e = 6/4 =3/2

For hyperbola,

e = √(1+ b²/a²) = 3/2

9/4 = 1 + b²/16

b²/16 = 5/4

b² = 20 

x²/16 – y²/20 = 1

5x² – 4 y² = 80.

Therefore, equation will be 5x² – 4y² = 80.            

Problem 4: If the locus of a hyperbola be 8 and its eccentricity is 3/√5. Find the equation of hyperbola?

Solution:  

Eccentricity (e) = 3/√5                               

b²/a² = 4/5 ⇢ (1)

Latus rectum = 8                   

2(b²)/a = 8

b²/a = 4 ⇢ (2)

Dividing (2) by (1)

(b²/a² )/(b²/a) = (4/5)/4

b² = 20  

So, a² = 25 ,b²=20

Equation of hyperbola:

x²/25 – y²/20 = 1

(4x² – 5y²)/100 = 1

4x² – 5y² = 100

Therefore equation will be 4x²– 5y² = 100

Problem 5: Locus of the feet of the perpendiculars drawn from either focus on a variable tangent to the hyperbola 16y² – 9x² = 1?

Solution:       

16y² – 9x² = 1

9x² – 16y² = -1

x²/(1/339) – y²/(1/16) = -1 

For conjugate hyperbola equation of auxiliary circle is: x² + y² = b²

So, b² = 1/16

Equation will be x² + y² = 1/16

Problem 6: Find the length of traverse axis of the hyperbola, when the product of the perpendicular distance from any point on the hyperbola x²/a²  – y²/b² = 1 having eccentricity e = √3 on its asymptotes is equal to 6.

Solution: 

Eccentricity (e)² = (a²/b² ) + 1

(√3)²  = (a²/b²) + 1

3 = (a²/b²) + 1

(a²/b²) = 2

(a²) = 2b²

Product of perpendicular = (a²b²)/a² +b² 

[2b² × b²]/(2b² + b²) = 6

[2b² × b²]/(3b²) = 6

2b² = 18

b² = 9

b = ± 3 

Therefore Length = 2 × 3 = 6

Frequently Asked Questions

How to minimize the function?

Minimum value of any function is found by manipulating the variables in the function and solving for the point where the function reaches its lowest value.

What does “minimize a function” mean?

Minimize a function means finding the lowest value of the funtionin any given interval, it is also known as infimum.

How to maximize and minimize any function?

We maximize or minimize any function by using the concept of derivatives.