A tangent to the curve y = 1 − x 2 is drawn so that the abscissa x 0 of the point of tangency belongs to the interval (0, 1]. The tangent at x 0 meets the x − axis and y − axis at A & B respectively. Then find the minimum area of the triangle OAB, where O is the origin
Text Solution
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( 
)
y = 1 – x 2
Consider point P (x 0 , 1 – 
)
0 < x 0 ≤ 1
equation of tangent at P is
y – (1 – 
) = – 2x 0 (x – x 0 )
intersection with x-axis at
⇒ A ≡ 
intersection with y-axis at
B(0, 2 
+ (1 – 
)
area of Δ OAB Δ = 
= 
= 
[3 
– 1] = 0 ⇒ x 0 = 
changes sign from +ve to – ve
at x 0 = 
So point of minimum ⇒ A min = 
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