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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

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

Grade:11

2 Answers

Vaibhav
50 Points
5 years ago
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
ankit singh
askIITians Faculty 614 Points
2 years ago

 

Let PQ be the normal chord to the parabola y2=4x where a=1
Let (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 is
y−2t1​=2−2t1​​(x−t12​)
⇒y−2t1​=−t1​(x−t12​) …………(1)
Since Q is a point at the normal, substituting y=2t2​ and x=t22​
we 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
⇒t12​2t1​​⋅t22​2t2​​=−1
⇒t1​t2​4​=−1
⇒t1​t2​=−4 ……(2)
From (2),
−t1​(t2​+t1​)=2
⇒−t1​t2​−t12​=2
⇒4−t12​=2 [from (3)]
⇒t12​=2
⇒t1​=2​
Substituting value of t1​ in (3) we have
t2​=2​−4​
t2​=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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