The shortest distance between the parabolas y^2 = x-1 and x^2= y-1 is 3?2/k . Then find the value of k .
Please do this question both by Maxima and minima and simple method . Please please please do it by both the methods . Do not forget to do it by both the methods .

Dear Student,
Please find below the solution to the asked query:

Key point: Shortest is obtained is obtained along common normalWe havey2=x-1....ix2=y-1...iiy2=x-1x=y2+1If we replace y and x. then we gety=x2+1x2=y-1 which is iiHence given parabolas are inverse  of each other.Hence they are mirror images of each other with respect to line y=x.

Slope of line y=x is m=1Now required point should have this slope of 1 for its tangent atpoint of tangency at ends of common normal.x2=y-1Differentiate with respect to x:2x=dydxSet dydx=Slope=12x=1x=12Put in iix2=y-114=y-1y=54Hence we have P12,54 as required point on iiAs i and ii are inverse of each other, henceQ54,12 is the required point on iHence required distance is given by:PQ=12-542+54-122=342+342=324According to question:324=32kk=4Note: I am not sure which simple method you are mentioning. Please belittle more specific. Although method Ihave adopted is the easiest approach.

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