If tangents are drawn at the points A(1,2),B(4,−4) and a variable point ′C′ lies on the parabola y2=4x, then the locus of the orthocentre of triangle formed by these tangents is
A
x=0
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B
x−1=0
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C
x+1=0
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D
y=0
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Solution
The correct option is Cx+1=0 For the given parabola a=1
Now, A≡(1,2)≡(t21,2t1)⇒t1=1B≡(4,−4)≡(t22,2t2)⇒t2=−2
Let C≡(t2,2t)
Intersection point of tangents at A and B =(t1t2,(t1+t2))≡(−2,−1)
Let P≡(−2,−1)
Intersection point of tangents at C and B =(−2t,t−2)
Let R≡(−2t,t−2)
Intersection point of tangents at C and A =(t,t+1)
Let Q≡(t,t+1)
Slope of QR=1t
Slope of PQ=1
Hence, the equation of altitiude through R on PQ is y+2−t=−(x+2t)⇒x+y+t+2=0⋯(1)
The equation of altitiude through P on QR is y+1=−t(x+2)⋯(2)
Therefore, the orthocentre is the point of intersection of equation (1) and (2) −x−t−2=−tx−2t−1⇒x(t−1)+t−1=0⇒(x+1)(t−1)=0⇒x+1=0(∵t≠1)
∴ The locus of orthocentre is x+1=0
Alternate Solution:
For any triangle formed by the tangents drawn at t1,t2,t3 on the parabola y2=4ax it's orthocentre will always lies on directrix, i.e. x+a=0
Hence, the locus is x+1=0