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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
4 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
11 months 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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