Three normals are drawn from the point (c, 0) to the curve y2 = x. Sho...
Proof that c must be greater than 1/2
Let (a, b) be any point on the curve y2 = x. Then the slope of the tangent at this point is given by:
dy/dx = 1/(2b)
Therefore, the slope of the normal at this point is given by:
m = -2b
Now, let (c, 0) be the point from which the normals are drawn. The equation of the normal passing through (a, b) is given by:
y - b = (-2b)(x - a)
This normal will pass through (c, 0). Substituting these values, we get:
0 - b = (-2b)(c - a)
or, c - a = 1/2
Therefore, c must be greater than 1/2 for there to exist a normal from (c, 0) to the curve y2 = x.
For what value of ‘c’ are the other two normals perpendicular to each other?
Let (a, b) and (p, q) be two points on the curve y2 = x such that the normals passing through (a, b) and (p, q) are perpendicular to each other. Let the point (c, 0) be the point from which the normals are drawn. The equations of the normals passing through (a, b) and (p, q) are given by:
y - b = (-2b)(x - a)
and
y - q = (-2q)(x - p)
respectively.
These two normals are perpendicular to each other. Therefore, their slopes must satisfy:
m1 × m2 = -1
where m1 and m2 are the slopes of the two normals.
Substituting the slopes, we get:
(-2b) × (-2q) = -1
or, bq = 1/4
Now, the two normals will pass through the point (c, 0). Substituting these values in the equations of the normals, we get:
y = -2bx + 2ab + b
and
y = -2qx + 2pq + q
respectively.
These two lines will be perpendicular to each other if and only if their slopes satisfy:
m1 × m2 = -1
Substituting the slopes, we get:
-2b × -2q = -1
or, bq = 1/4
This is the same condition that we obtained earlier. Therefore, for the two normals passing through (a, b) and (p, q) to be perpendicular to each other, the value of c must satisfy:
bq = 1/4
Substituting b