The equation of the tangent at the vertex of the parabola x2 + 4x + 2y = 0, is
x = - 2
x = 2
y = - 2
y = 2
The equation of the normal to the hyperbola x2 - 16y2 - 2x - 64y - 72 = 0 at the point (- 4, - 3) is
5x + 16y + 79 = 0
16x + 5y + 97 = 0
16x + 5y + 79 = 0
5x + 16y + 97 = 0
C.
16x + 5y + 79 = 0
First, we find the equation of tangent to the hyperbola
x2 - 16y2 - 2x - 64y - 72 = 0 ...(i)
at the point (-4,- 3), which is given by
4x + 48y - (x - 4) - 32 (y - 3) - 72 = 0
- 4x + 48y - x + 4 - 32y + 96 - 72 = 0
- 5x + 16y + 28 = 0 ...(ii)
Required normal is the normal to the tangent (ii), which is given by
16x + 5y + c = 0 ...(iii)
But it passes through (- 4, - 3).
Putting this value in Eq. (iii), we get
5x + 16y + 79 = 0
Which is the required normal.
The centre of the circle which circumscribes the square formed by x2 - 8x + 12 = 0 and y2 - 14y + 45 = 0 is
(4, 5)
(3, 4)
(9, 5)
(4, 7)
The point of contact of 3x + 4y + 7 = 0 and x2 + y2 - 4x - 6y -12 = 0 is
(1, 1)
(- 1, 1)
(1, - 1)
( -1, - 1)
The equation x2 - 7xy + 12y2 = 0 represents a
circle
pair of parallel straight lines
pair of perpendicular straight lines
pair of non-perpendicular straight lines
The equation of circle passing through the points (0, 2) (3, 3) and having its centre on the x-axis is
x2 + y2 - 14x - 12 = 0
3x2 + 3y2 - 22x - 4 = 0
3x2 + 3y2 - 14x - 12 = 0
None of the above