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Question

# Tangent to the circle x2+y2 = 4 at any point on it in the first quadrant makes intercepts OA and OB on x and y axes respectively, O being the centre of circle. Find the minimum value of (OA + OB).

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Solution

## Given circle x2+y2=4 ....(i) Let P (α,β) ∴α2+β2 ....(ii) Also 2x+2ydydx=0 ⇒dydx=−xy ∴dydx∣∣at P=−αβ=mr Equation of tangent : y−β=−αβ(x−a) ⇒αx+βy=a2+β2 By (ii), αx+βy=4 ⇒x4α+y4β=1 ∴x−intercept=4α=OA,y−intercept=4β=OB Let OA+OB=s=4α+4β ⇒s=4α+4√4−α2 ⇒dsdα=−4α2+4α(4−α2)32 ⇒d2sdα2=8α3+(4−α2)324−4α×32(4−α2)12(−2α)(4−α2)3 ⇒d2sdα2=8α3+8(α2+2)(4−α2)52 For local maxima and/or minima, dsdα=−4α2+4α(4−α2)32=0 ⇒α3=(4−α2)32 ⇒α6=(4−α2)3=64−48α2+12α4−α6 ⇒m3−6m2+24m−32=0, where m = α2 ⇒(m−2)=0 or (m2−4m+16)=0 ∴m=2,m2−4m+16≠ 0 ∀ m ϵ R ∴α=√2. [∴ P lies in 1st quadrant] ∴d2sdα2at α=√2∣∣=82√2+8(2+2)(4−2)52=6√2>0 So, s is minimum at α=√2. Therefore, the minimum value of s =4√2+4√4−2=4√2.

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