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Question: Let (tan a) x + (sin a) y = a and (a cosec a)x + (cosa) y = 1 be two straight lines, a being the par...

Let (tan a) x + (sin a) y = a and (a cosec a)x + (cosa) y = 1 be two straight lines, a being the parameter. Let P be the point of intersection of the lines. In the limiting position when a ® 0 the coordinate of P are

A

(2, 1)

B

(2, –1)

C

(–2, 1)

D

(–2, – 1)

Answer

(2, –1)

Explanation

Solution

Solve the equation for values of x and y

x = αcos⁡α−sin⁡αsin⁡α−α\frac { \alpha \cos \alpha - \sin \alpha } { \sin \alpha - \alpha } y = α−xtan⁡αsin⁡α\frac { \alpha - x \tan \alpha } { \sin \alpha }

where x = lim⁡α→0\lim _ { \alpha \rightarrow 0 } αcos⁡α−sin⁡αsin⁡α−α\frac { \alpha \cos \alpha - \sin \alpha } { \sin \alpha - \alpha } (0/0)

L.Hospital

x = −αsin⁡α+cos⁡α−cos⁡αcos⁡α−1\frac { - \alpha \sin \alpha + \cos \alpha - \cos \alpha } { \cos \alpha - 1 }

x = lim⁡α→0\lim _ { \alpha \rightarrow 0 }– αsin⁡αcos⁡α−1\frac { \alpha \sin \alpha } { \cos \alpha - 1 } (0/0)

x = lim⁡α→0\lim _ { \alpha \rightarrow 0 } α(2sin⁡α2cos⁡α2)2sin⁡2α/2\frac { \alpha \left( 2 \sin \frac { \alpha } { 2 } \cos \frac { \alpha } { 2 } \right) } { 2 \sin ^ { 2 } \alpha / 2 }

x = αtan⁡α2\frac { \alpha } { \tan \frac { \alpha } { 2 } } = lim⁡α→0\lim _ { \alpha \rightarrow 0 } (α2)×2tan⁡(α2)\frac { \left( \frac { \alpha } { 2 } \right) \times 2 } { \tan \left( \frac { \alpha } { 2 } \right) }

= + 2

and y = lim⁡α→0\lim _ { \alpha \rightarrow 0 } lim⁡α→0\lim _ { \alpha \rightarrow 0 } tan⁡αsin⁡α\frac { \tan \alpha } { \sin \alpha } × x

y Ž lim⁡α→0\lim _ { \alpha \rightarrow 0 } lim⁡α→0\lim _ { \alpha \rightarrow 0 }

{lim⁡α→0x=+2}\left\{ \lim _ { \alpha \rightarrow 0 } x = + 2 \right\}

y = 1 – 2 = – 1 Hence (2, – 1)