Equation of Parabola When Its Axis Is Parallel to X or Y Axis
Circles are d...
Question
Circles are drawn taking any two focal chords of the parabola y2=4ax as diameters. If t1,t3 represent any one end of the different focal chords, then, the equation of the common chord of the circles is
A
x(t3+t1)(t21t23−1)+2yt1t3(t1t3+1)=0
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B
x(t3+t1)(t21t23+1)=2yt1t3(t1t3+1)
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C
x(t3−t1)(t21t23+1)=2yt1t3(t1t3+1)
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D
x(t3−t1)(t21t23+1)+2yt1t3(t1t3+1)=0
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Solution
The correct option is Ax(t3+t1)(t21t23−1)+2yt1t3(t1t3+1)=0 Let PQ and RS be two focal chord of the given parabola and let P,Q,R,S be the points t1,t2,t3,t4 respectively, then t1t2=−1 and t3t4=−1 As per the question PQ is a diameter for a perticular circle, then the equation of the circle is (x−at21)(x−at22)+(y−2at1)(y−2at2)=0⇒x2+y2−a(t21+t22)x−2a(t1+t2)y−3a2=0…(1) and circle with the diameter as RS will be x2+y2−a(t23+t24)x−2a(t3+t4)y−3a2=0…(2) Hence the equation of common chord will be ax(t23+t24−t21−t22)+2ay(t3+t4−t1−t2)=0⇒ax((t23−t21)+1t23−1t21)+2ay((t3−t1)+1t1−1t3)=0⇒ax(t23−t21)(t21t23−1)+2ay(t3−t1)t1t3(t1t3+1)=0 as t1≠t3 ∴x(t3+t1)(t21t23−1)+2yt1t3(t1t3+1)=0