If the vertices of a· triangle are A(0, 4, 1), B(2, 3, - 1) and C(4, 5, 0), then the orthocentre of ABC, is
(4, 5, 0)
(2, 3, - 1)
(- 2, 3, - 1)
(2, 0, 2)
Two lines and intersect at a point, if k is equal to
C.
x = 2r + 1, y = 3r - 1, z = 4r + 1
Since, the two lines intersect.
So, putting above values in second line, we get
Taking 1st and 3rd terms, we get
2r - 2 = 4r + 1
r = - 3/2
Also, taking 2nd and 3rd terms, we get
3r - 1 - k = 8r + 2
k = - 5r - 3 = - 3 =
The normals at three points· P,Q and R of the parabola y2 = 4ax meet at (h, k). The centroid of the PQR lies on
x = 0
y = 0
x = - a
y = a
The minimum area of the triangle formed by any tangent to the ellipse ( x2/a2 ) + ( y2/b2 ) = 1 with the coordinate axes is
a2 + b2
( a + b )2/2
ab
( a - b )2/2
If the line lx + my - n = 0 will be a normal to the hyperbola, then , where k is equal to
n
n2
n3
None of these