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If `(x-a)^2+(y-b)^2=c^2`, for some `c > 0`, prove that `([1+((dy)/(dx))^2]^(3/2))/((d^2y)/(dx^2))`is a constant independent of a and b. |
Answer» `(x-a)^(2)+(y-b)^(2)=c^(2) " "` …(1) `implies2(x-a)+2(y-b)(dy)/(dx)=0` `implies(dy)/(dx)= -(x-a)/(y-b) " " ` ...(2) `implies (d^(2)y)/(dx^(2))= -((y-b)*1-(x-a)(dy)/(dx))/((y-b)^(2))` `= -((y-b)-(x-a)[-(x-a)/(y-b)])/((y-b)^(2))` `= -((y-b)^(2)-(x-a)^(2))/((y-b)^(3))` `= -(c^(2))/((y-b)^(3))` ` " " `From equation (1) Now `([1+((dy)/(dx))^(2)]^(3/2))/((d^(2)y)/(dx^(2)))=([1+((x-a)/(y-b))^(2)]^(3/2))/((-c^(2))/((y-b)^(3)))` `=([((y-b)^(2)+(x-a)^(2))/((y-b)^(2))]^(3//2))/(-(c^(2))/((y-b)^(3)))` `= -[(c^(2))/((y-b)^(2))]^(3//2)*((y-b)^(3))/(c^(2))` `= -(c^(3))/((y-b)^(3))*((y-b)^(3))/(c^(2))= -c` which is a constant independent of a and b. ` " " `Hence Proved. |
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