Let, #P=P(h,k)# be an arbitrary point on line #l : x-y+3=0#.
#"As, "P in l, h-k+3=0:. k=h+3 rArr P=P(h,h+3)#.
Recall that the eqn. of tangent at the point #(x_0,y_0)# on the
Parabola # S : y^2=8x# is given by, # yy_0=4x+4x_0#.
#(x_0,y_0) in S rArr y_0^2=8x_0, or, x_0=y_0^2/8," so that, "#
# yy_0=4x+4y_0^2/8, i.e., 2yy_0=8x+y_0^2............(star^1)#.
Suppose that, the tangents #t_1, & t_2#, drawn from #P in l" to "S#
touch #S# at the points #Q(x_1,y_1) and R(x_2,y_2)#, resp.
#(star^1)rArr t_1 : 2yy_1=8x+y_1^2; t_2 : 2yy_2=8x+y_2^2#.
But, #P in t_1, and, P in t_2#.
#:. 2(h+3)y_1=8h+y_1^2, and, 2(h+3)y_2=8h+y_2^2#.
These eqns. suggest that, #y_1 and y_2# are the roots of the
quadr. eqn. # y^2-2(h+3)y+8h=0#.
#:. y_1+y_2=2(h+3), and y_1y_2=8h............(star^2)#.
Our task is to determine the fixed point through which the chord
#QR# passes through.
The chord #QR : |(x,y,1),(x_1,y_1,1),(x_2,y_2,1)|=0, i.e.,#,
#x(y_1-y_2)-y(x_1-x_2)+(x_1y_2-x_2y_1)=0#.
But, #Q,R in S :. y_1^2=8x_1, y_2^2=8x_2:. x_1=y_1^2/8, x_2=y_2^2/8#.
#:. QR : x(y_1-y_2)-y/8(y_1^2-y_2^2)+(y_1^2/8*y_2-y_2^2/8*y_1)=0#
#:. 8x(y_1-y_2)-y(y_1-y_2)(y_1+y_2)+y_1y_2(y_1-y_2)=0#.
#:. QR : 8x-y(y_1+y_2)+y_1y_2=0...[because, y_1ney_2]#.
Utilising #(star^2)# in this eqn. we get,
# QR : 8x-2(h+3)y+8h=0, or, 4x-(h+3)y+4h=0#.
#:. QR : (4x-3y)+h(4-y)=0#.
Recall that, # AA lambda in RR, (ax+by+c)+lambda(lx+my+n)=0#
represents a line passing through the point of intersection
of the lines #ax+by+c=0, and lx+my+n=0#.
Accordingly, we conclude that #QR# passes through,
#"the point "in {(x,y) | 4x-3y=0}nn{(x,y) | y-4=0}#.
Evidently, #(3,4)# is the desired fixed point!
Enjoy Maths., and Spread he Joy!
