Ajay Verma
Last Activity: 11 Years ago
solution:
for any curve the equation of normal at ( X1, Y1)
(y- y1) = -1/m (x- X1) {where m = slope of tengent or say dy/dx at (x1, Y1) }
normal meets the X axis at G..
so
( 0 - y1) = -1/m (x- X1)
so x = m*Y1 + X1
so coordinates of G( m*Y1 + X1 , 0)
now ...distance of G from the origin is twice the abscissa of P
so
2X1 = m*Y1 + X1
X1 = m*Y1
m = X1/Y1
dy/dx (at X1, Y1) = X1/Y1
dy/dx = x/y
ydy = x dx
integrate both sides
y2/2 = x2/2 + c
x2/2 - y2/2 = c
so hyperbola..
Thanks and Regards,
Ajay verma,
askIITians faculty,
IIT HYDERABAD