Maths-
General
Easy
Question
PA and PB are the tangents drawn to y2 = 4x from point P. These tangents meet the y-axis at the points A1 and B1 respectively. If the area of triangle PA1 B1 is 2 sq. units, then locus of ‘p’ is -
- a2 (y2 + 2x)x2 = 8
- a2 (y2 + 4x)x2 = 16
- a2 (y2 – 4x)x2 = 8
- a2 (y2 – 4x)x2 = 16
The correct answer is: a2 (y2 – 4x)x2 = 16
Let A ≡ (t12, 2t1), B ≡ (t22, 2t2), then P ≡ (t1t2, t1 + t2).
Also equations of PA and PB are yt1 = x + at12, yt2 = x + at22.
Thus, A1 ≡ (0, at1), B1 ≡ (0, at2) .
Now, area of triangle PA1B1 = . |A1B1| |t1 t2|
= a |t1 – t2| |t1 t2| = 2(given)
a2 (t1 – t2)2 (t1t2)2 = 16
a2 [(t1 + t2)2 – 4t1t2] (t1t2)2 = 16
Thus locus of ‘P’ is a2 (y2 – 4x) x2 = 16.
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