PA and PB are the tangents drawn to y 2 = 4x from point P. These tangents meet the y-axis at the points A 1 and B 1 respectively. If the area of triangle PA 1 B 1 is 2 sq. units, then locus of ‘p’ is-
Text Solution
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Let A ≡ (t 1 2 , 2t 1 ), B ≡ (t 2 2 , 2t 2 ), then P ≡ (t 1 t 2 , t 1 + t 2 ).
Also equations of PA and PB are yt 1 = x + at 1 2 , yt 2 = x + at 2 2 .
Thus, A 1 ≡ (0, at 1 ), B 1 ≡ (0, at 2 ) .
Now, area of triangle PA 1 B 1 = 
. |A 1 B 1 | |t 1 t 2 |
= 
a |t 1 – t 2 | |t 1 t 2 | = 2 (given)
⇒ a 2 (t 1 – t 2 ) 2 (t 1 t 2 ) 2 = 16
⇒ a 2 [(t 1 + t 2 ) 2 – 4t 1 t 2 ] (t 1 t 2 ) 2 = 16
Thus locus of ‘P’ is a 2 (y 2 – 4x) x 2 = 16.
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