Application of DerivativeHard
Question
The equation of the common tangent to the curves y2 = 8x and xy = - 1 is
Options
A.3y = 9x + 2
B.y = 2x +1
C.2y = x + 8
D.y = x + 2
Solution
Tangent to the curve y2 = 8x is y = 
. So it must satisfy xy = -1.
x
= - 1 ⇒ 
x + 1 = 0
Since it has equal roots, therefore D = 0
⇒
- 4m = 0
⇒ m3 = 1 ⇒ m = 1
So equation of common tangent is y = x + 2
. So it must satisfy xy = -1.x
= - 1 ⇒ x + 1 = 0Since it has equal roots, therefore D = 0
⇒
- 4m = 0⇒ m3 = 1 ⇒ m = 1
So equation of common tangent is y = x + 2
Create a free account to view solution
View Solution FreeMore Application of Derivative Questions
The slope of tangent to a curve y = f (x) in [x, f (x)] is 2x + 1 If the curve passes through the point (1, 2) then the ...The set of values of p for which the extremum of the function f(x) = x3 - 3 px2 + 3 (p2 - 1) x + 1 lie in the interval (...The chord joining the points where x = p and x = q on the curve y = ax2 + bx + c is parallel to the tangent at the point...The equation of the normal to the curve 2y = 3 - x2 at (1, 1) is-...If tn be the nth term of the series 1 + 3 + 7 + 15 + ......... then -...