chalte chalte good to know

hey come on somone

For a parabola at any point (at², 2at) slope of normal is given by -t

=> m= -t

we can write the point as (am², -2am)

if this point lies on the given line then :

-2am = am³+ c

Hopefully this is correct

hey these are useful for quick revision!!!

For hyperbola, the equation of normal at any point(asec θ, btanθ) is-

axsinθ+ by= (a²+b²)tanθ

equation of the line is y-mx=c

These would represent the same line if

asinθ/-m = b = (a²+b²)tanθ/c
→ cosecθ=-a/mb and cotθ= a²+b²/bc

cosec²Î¸-cot²Î¸=1

Solving,
a²c²=m²[b²c²+(a²+b²)²]

For ellipse,
any equation is ax secθ- by cosecθ=a²-b²
and line is y-mx=c

Solve it in the same way as in for hyperbola to get the result..