If θ be the angle subtended at the focus by the chord which is normal at the point (λ,λ),λ≠0 to the parabola y2=4x, then the equation of the line making angle θ with positive x−axis and passing through (1,2) is
A
y=2
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B
x+2y=5
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C
x+y=3
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D
x=1
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Solution
The correct option is Dx=1 Let P≡(t21,2t1) and Q≡(t22,2t2)
Putting (λ,λ) in y2=4x, we get λ=0,4
But λ≠0 (Given)
So,λ=4 ∴P≡(4,4)⇒t1=2
Also, t2=−t1−2t1
Hence, t2=−3
⇒Q≡(9,−6)
Now, Slope of PF,m1=43
and Slope of QF,m2=−34 m1m2=−1 ∴θ=90∘
Hence, equation of line passing through (1,2) and inclined at angle θ=90∘ is x=1