a little tricky question

Calculus Level 3

What are the equations of the mutual tangent lines to y = x 2 6 x + 10 y=x^{2} - 6x +10 and y = 4 x x 2 7 y=4x -x^{2} - 7 ?

there is no mutual tangential 4 x y + 4 = 0 4x-y+4=0 and x 3 y + 7 = 0 x-3y+7=0 4 x + y 9 = 0 4x+y-9=0 and 2 x y 6 = 0 2x-y-6=0 x 3 y + 15 = 0 x-3y+15=0 and x + 3 y 18 = 0 x+3y -18=0

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1 solution

Ahmed Mahmoud
Apr 7, 2018

lets find the intersection points of the two linear equations x^2 -6x +10=4x- x^2-7 no real solutions the two curves don't intersect lets suppose that the tangent intersects the first curve at (A , B) and the second curve at (C, D) the slope of the tangent line would be (D-B)/(C-A)
and this is equal to (4C-C^2-7)-(A^2-6A+10) / (C-A) (1) the slope of the first tangent is equal to ∆y⁄∆x =(2A-6) (2) the slope of the second tangent is equal to ∆y⁄∆x =(4-2C) (3) solving (2) and (3) together so (A+C=5) (4) FROM (1) ,(2),(3) (4C-C^2-7)-(A^2-6A+10)/(C-A) =(2A-6) FROM (4) SO (20-4A-25+10A-A^2-7-A^2+6A-10)=(-4A^2+22A-30) (A^2-5A+4=0) so ((A-1)(A-4)=0) IF A=1 SO B =5 so the point is (1,5) so the slope =-4 so the equation of the tangent is 4x+y-9=0 OR A=4 SO B=2 SO THE POINT IS (4,2) SO THE SLOPE =2 so the equation of the tangent is 2x-y-6=0

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