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Length of a normal chord of the parabola,y^2=4x, which makes an angle of pi/4 with the axis of x is.Answer is 8 root 2 :( please help its urgent
Eq of normal y=mx-am^3-2 am since m=1 a=1 putting value give y=x-3 put y=x-3 in y^2=4x solving gives x=1,9 from y=x-3 get y=-2 , 6 points known then from distance formula you get answers please approved if understand
Let PQ be the normal chord to the parabola y2=4x where a=1Let (at12,2at1) and (at22,2at2) be the coordinates of P and Q respectively.i.e., coordinates of P=(t12,2t1)and coordinates of Q=(t22,2t2)Equation of normal at point P isy−2t1=2−2t1(x−t12)⇒y−2t1=−t1(x−t12) …………(1)Since Q is a point at the normal, substituting y=2t2 and x=t22we have,2t2−2t1=−t1(t22−t12)⇒2(t2−t1)=−t1(t2+t1)(t2−t1)⇒2=−t1(t2+t1)……… (2)It is given that the normal chord subtends a right angle at the vertex A(0,0) i.e., PA⊥AQ∴ (slope of PA)(slope of AQ)=−1((t12−0)(2t1−0))(t22−02t2−0)=−1⇒t122t1⋅t222t2=−1⇒t1t24=−1⇒t1t2=−4 ……(2)From (2),−t1(t2+t1)=2⇒−t1t2−t12=2⇒4−t12=2 [from (3)]⇒t12=2⇒t1=2Substituting value of t1 in (3) we havet2=2−4t2=2−42=−22∴ coordinates of P=((2)2,2(2))=(2,22)Coordinates of Q=((−22)2,2(−22))=(−8,−42)∴ Length of normal chord PQ=(8−2)2+(−42−22)2=62+(−62)2=36+72=108=63.
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