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The equation of the parabola to which the line m^2(y-10)-mx-1=0 is a tangent for any real value of m
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hello mounika. fromhttps://www.askiitians.com/iit-jee-coordinate-geometry/tangent-to-a-parabola.aspxwe know that am^2) for all values of m.2am, at (= 4ay is a tagnent to the parabola x^2y = mx - am^2 ….... (1)m^2(y-10)-mx-1=0 can be written as (y-10)= x/m+1/m^2 or Y= nx – ( – 1)n^2 .........(2), where n= 1/m (and hence n can also be any real since m can vary over the reals) and Y= y – 10on comparing (2) and (1) we have a= – 1so, the corresponding parabola would be x^2 = 4aY substituting valuesx^2 = 4*( – 1)(y – 10)or x^2 + 4(y – 10) = 0kindly approve :))
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