Find the angle of  intersection of 2 curve x^2y=2 and xy^2=4

 

 

Asked by agarwalgolu318 | 13th Aug, 2020, 08:09: PM

Expert Answer:

The given curves are x2y = 2 ... (i) and xy2 = 4 ... (ii)
From (i), we get y=2/x2
Substituting in (ii), we get x =1
So, we get y = 2
Thus, the point of intesec of the two curves is (1, 2)
Let space straight m subscript 1 space and space straight m subscript 2 space be space the space slopes space of space curves space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space respectively
From space left parenthesis straight i right parenthesis space rightwards double arrow space dy over dx equals negative 4 over straight x cubed rightwards double arrow straight m subscript 1 equals right enclose dy over dx end enclose subscript left parenthesis 1 comma space 2 right parenthesis end subscript equals negative 4
From space left parenthesis ii right parenthesis space rightwards double arrow space dy over dx equals negative fraction numerator 2 over denominator straight x squared straight y end fraction rightwards double arrow straight m subscript 2 equals right enclose dy over dx end enclose subscript left parenthesis 1 comma space 2 right parenthesis end subscript equals negative 1
Let space straight theta space be space the space angle space between space the space two space curves
Angle space between space the space curves space is space given space by colon
tanθ equals open vertical bar fraction numerator straight m subscript 1 minus straight m subscript 2 over denominator 1 plus straight m subscript 1 straight m subscript 2 end fraction close vertical bar equals open vertical bar fraction numerator negative 4 plus 1 over denominator 1 plus 4 end fraction close vertical bar equals 3 over 5
rightwards double arrow straight theta equals tan to the power of negative 1 end exponent open parentheses 3 over 5 close parentheses

Answered by Renu Varma | 14th Aug, 2020, 11:03: AM