If the normal at the point P(at2 , 2at) to the parabola y2=4ax cuts the circle drawn with PS as diameter at Q ( S- focus ) P.T. PQ = a√( 1+t2)
1Please suggest a shortcut , do we have no other way than writing the normal and finding it's intersection point?
62you can solve it by geometry alone
take the angle between normal and diameter PS
Also the radius of the circle is simple to find...
I hope this is hint enuf?
1I got the radius a(1+t2)/2 , then the angle between the normal and the radius is tanθ = t . then , the side opposite θ is tx and adjacent is x . So we ,get a√(1+t2) / 2 . thus the length is a√(1+t2)
Thank you!!!!!!!! Nishant . But just check if I've made any mistakes .You're an awesome teacher.
62Thanks buddy :)
hey yeah there is a slight mistake.. u need to take diameter .. not radius :)
the right angled traingle has hypotenuse as D :)
62hey there is another brilliant way to do this..
find the distance between the line through the focus and parallel to the tangent at P(at2 , 2at)
does this make sense?
1or take radius and then , find half of the chord , we know θ and hypotenuse . perpendicular from center to chord bisects it.then we can double it. SAME , isn't it?
Thanks buddy :)
