find the equation of tangents drawn to the curve y2 - 2x3 -4y + 8 from the point (1,2) ????
1the slope of any tangent = (2 + 3x)/y.
Now let the point on the curve be (x1,y1).
The slope of the tangent is also (y1-2)/(x1 - 1) = (2 + 31)/y1.
Equating, you will get an equation. Solve that equation and the curve to get the 2 points where the tangent and the curve meet.
Then get the eq of tangent,,
1varun tel me your final answer and how u get slope
i think slope should be 3 x2 /(y-2)
1My bad ... I made a mistake in slope :p
The slope = 3x2/(y-2).
Let the point be (x,y).
(y-2)/(x-1) = 3x2/(y-2).
=> 3x3-3x2-y2+4y-4 = 0
Now eq of curve =
y2-2x3-4y+8 = 0
1357slope = 3x2/(y-2)
let the point be (x1,y1)
eqn of tangent = (y-2)/(x-1)=3x12/(y1-2)
as it passes through (x1,y1)
the eqn is
(y1-2)/(x1-1)=3x12/(y1-2)
i.e. (y1-2)2=3x12(x1-1)
also given curve is (y1-2)2=2x13-4
from two eqns we get
2x13-4=3x12(x1-1)
i.e x13-3x12+4=0
(x1+1)(x1-2)2=0
so x1=2 (x1=-1 not possible)
and y1=2±2√3