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Question: The minimum length of radius vector of the curve \(\frac{a^{2}}{x^{2}} + \frac{b^{2}}{y^{2}}\)= 1 i...

The minimum length of radius vector of the curve

a2x2+b2y2\frac{a^{2}}{x^{2}} + \frac{b^{2}}{y^{2}}= 1 is

A

|a – b|

B

|a + b|

C

2|a + b|

D

None of these

Answer

|a + b|

Explanation

Solution

a2r2cos⁡2φ+b2r2sin⁡2φ\frac{a^{2}}{r^{2}\cos^{2}\varphi} + \frac{b^{2}}{r^{2}\sin^{2}\varphi}= 1 (Let x = r cosf, y = r sinf)

r2 = a2sec2f + b2cosec2f

Let r2 = z, z = a2sec2f + b2cosec2f

dzdφ\frac{dz}{d\varphi}= a22 secf secf tanf – b2 2 cosec f cosec f cot f = 0

a2 sec2 f tan f = b2 cosec2 f cot f

tan4 f = b2a2\frac{b^{2}}{a^{2}} Ž tan f = ba\sqrt{\frac{b}{a}} or −ba- \sqrt{\frac{b}{a}}

rmin⁡2r_{\min}^{2}= a2(1+ba)\left( 1 + \frac{b}{a} \right)+ b2(1+ab)\left( 1 + \frac{a}{b} \right)= (a + b)2

\ rmin = |a + b|