Question

if M and m are the minimum and maximum values of y/x for pair of real numbers (b,y) which satisfies the equation ( x-3)² +(y-3)²=6 find the value of m and m ​

Answer 1

thanks for points also sorry

Answer 2

Explanation:

Given if M and m are the minimum and maximum values of y/x for pair of real numbers (b,y) which satisfies the equation ( x-3)² +(y-3)²=6 find the value of 1/M and 1/m  

• Now equation of the circle (x – 3)^2 + (y – 3)^2 = 6
•      So centre = (3,3)
• Now we have sin theta = √6 / 3√2
•                         sin theta = 1/√2
• Now b = √(√3)^2 – (1)^2
•           = √3 – 11  
•          = √2
•      So tan theta = 1/√2
•     m + n = tan (45 + theta) + tan (45 – theta)
•                = 1 + tan theta / 1 – tan theta + 1 – tan theta / 1 + tan theta
•                = (1 + tan theta)^2 + (1 – tan theta)^2 / 1 – tan^2 theta
•                 = 2 + 2 tan theta / 1 – tan^2 theta
•                = 2 + 2 (1/√2)^2 / 1 – (1/√2)^2
•                 =  2 + 1 / 1 – ½  
•                  = 6
•    mn = (1 + tan theta / 1 – tan theta) (1 – tan theta / 1 + tan theta)  
•          = 1
•      Now 1/m + 1/n = m + n / mn
•                                 = 6/1
•                                    = 6
•      So we get 1/M + 1/m = 6

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https://brainly.in/question/16076442