CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If the tangents drawn to the parabola at the extremities of a common chord AB of the circle x2+y2=5 and the parabola y=bx2 intersect at the point T which lies on the directrix of the parabola, then 1b=
  1. 2
  2. 2
  3. 4
  4. 4


Solution

The correct options are
C 4
D 4

x2=yb
Comparing with x2=4ay
a=14b and Directrix :y=14b
Parametric coordinates of A is 
(2at,at2)=(t2b,t24b)

Since, A and B lie on the circle also,
coordinates of B is (t2b,t24b) 

y=bx2y=2bx
Slope at point A, yA=t
Slope at point B, yB=t

Equation of tangents at A and B are
(yt24b)=t(xt2b)     (1)
(yt24b)=t(x+t2b)     (2)

From (1) and 2, we get 
x=0, y=t24b
t24b=14bt=±1

 Therefore, points A and B are (±12b,14b)
A and B lie on the circle x2+y2=5
14b2+116b2=5
b=±14

b=14 (Upward parabola)
and b=14 (Downward parabola)

1b=±4

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image