If the straight line y - 2x+ 1 = 0 is the tangent to the curve xy + ax + by = 0 at x = i, then the values of a and b are respectively
1 and 2
1 and - 1
- 1 and 2
1 and - 2
D.
1 and - 2
Given line is y - 2x + 1 = 0 ...(i)
and curve xy + ax + by = 0 ...(ii)
Put x = 1 in Eq. (i), we get
y - 2(1) + 1 = 0 y = 1
From Eq. (i), we get
y = 2x - 1
Put y = 2x - 1 in Eq. (ii), we get
x (2x - 1) + ax + b(2x - 1) = 0
2x2 + (- 1 + a + 2b)x - b = 0
Since, Eq. (i) is a tangent to the Eq. (ii).
Discriminant, B2 - 4AC = 0
(- 1 + a + 2b)2 + 8b = 0 ...(iii)
Also, point (1, 1) satisfy the Eq. (ii)
1 1 + a + b = 0
a = - b - 1 ...(iv)
The equation of the tangent to the curve = 1 at the point (x1, y1) is . Then, the value of k is
2
1
3
3
The slope of the normal to the curve x = t2 + 3t - 8 and y = 2t2 - 2t - 5 at the point (2, - 1) is
-
If y = 4x - 5 is a tangent to the curve y = px3 + q at (2, 3), then (p + q) is equal to
- 5
5
- 9
9
The point on the curve y = 5 + x - x2 at which the normal makes equal intercepts is
(1, 5)
(0, - 1)
(- 1, 3)
(0, 5)