1.Tangents at the three points of a parabola y^2 = 4ax
form a equilateral triangle . Prove that the vertices of the triangle lie on the curve (3x + a)(3a + x) = y^2
2.Prove that the normal to the parabola y^2 = 4ax at (am^2, -2am) intersect the parabola at an angle \tan^{-1} (m/2)
3.If two tangents to the parabola y^2 = 4ax from a point P makes angle \alpha and \beta with the axis such that \tan^2\alpha + \tan^2\beta = \lambda ( constant)
then find the locus of P.
391
let the 3 vertickes be (at12,2at1)
(at22,2at2) , (at32,2at3)
now eqtan for a tangent throught (at^2,2at) is
yt=x+at^2
substitue those 3 points to get the eqtns of tangents and then get the point of intersection
u will find dat those points satisfy the equation of the cure ve u have give i,e (3x+a)(3a+x)=y2
392
eqtn for normal is y+xt=2at+at^3
put the point (am^2,-2am) in the above eqtn and u get the normal
now u have 2 eqtns with u
eqtn of the normal and eqtn of a curve
angle between these 2 is given by the angle between the tangent at the point where the normal intersects the parabola and the normal itself
393rd one
let the point be (x,y)
now find the equations of the 2 tangents that can be drawn from it to
the parabola y^2=4ax
then use the conditions given
thats all
