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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

Mounika , 5 Years ago
Grade 12
anser 2 Answers
Arun

Last Activity: 5 Years ago

Dear student
 
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Aditya Gupta

Last Activity: 5 Years ago

hello mounika. fromwe 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 – 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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