Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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
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.
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 :))
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -