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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Arun · Answer

Dear student Question is incomplete. Please check and repost the question. You can also attach an image. Na dif possible please post options too. as they are required.

Aditya Gupta · Answer

hello mounika. from https://www.askiitians.com/iit-jee-coordinate-geometry/tangent-to-a-parabola.aspx we know that am^2) for all values of m. 2am, at ( = 4ay is a tagnent to the parabola x^2 y = 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 – 10 on comparing (2) and (1) we have a= – 1 so, the corresponding parabola would be x^2 = 4aY substituting values x^2 = 4*( – 1)(y – 10) or x^2 + 4(y – 10) = 0 kindly approve :))

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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

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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