= 2a/3
Hence = 1/3. Hence e2 = 1- = 1 - (1/3) = 2/3
Therefore e =
The equation of the normal to a parabola y2 = 4ax is
y = mx -2am - am3
Hence the equation of the normal to the given parabola y2 = x is
mx - -
Since it passes throught (c,0)
mc -- = 0
on solving we get
m = 0 or m2 = 4(c-1/2)
If m = 0 then the equation of the normal is y = 0
If m2 0, then 4(c-1/2) 0
Hence c-1/2 0 or c > 1/2
s = (ae,0) and T = (-ae,0) and B = (0,b)
Since it is an equilateral triangle, ST2 = TB2
This implies 4a2e2 = a2e2 + b2
3a2e2 = b2
3a2e2 = a2(1 - e2)
3e2 = 1 - e2
Therefore e = 1/2
The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle is
Since the circle passes through (0,0) the equation reduces to
c= 0 -----(1)
Since it passes through (1,0),
1 + 2g + c = 0
This implies g = -1/2
Since the circle touches the circle x2 + y2 = 9, their radii should be equal
2 = 3
Substituting the values and simplifying we get f =
Hence the centre is (1/2, )
The given equation of the circle can be written as
(x-1)2 - 1 + (y- 2)2 - 4 = 0
(x - 1)2 + (y + 2)2 = 5
This implies the radius is and the centre is (1,-2)
The given point is (-1,2)
The distance between the centre of the circle and the given point is
=
Sice this is greater than the radius, the point lies outside the circle. Hence two tangents can be drawn.
slope form of normal is y=mx-2am-am3
for the given parabola a = 2 and m = 1
therefore y = x -4 -2
i.e; x - y - 6 = 0
comparing the above equation with we get center as (1,2)
putting (1,2) in L.H.S of above line ,we get 1+2=3=R.H.S
=> it satisfies center satisfies the equation of line or the line passes through center
and the line passing through the centre is normal.