Question

r=a2t by monge`s theorem

Answer 1

In geometry, Monge's theorem, named after Gaspard Monge, states that for any three circles in a plane, none of which is completely inside one of the others, the intersection points of each of the three pairs of external tangent lines are collinear.

Answer 2

Question:

By using Monge's Theorem solve r = a²t .

Answer:

r = a²t   is solved by using Monge's method.

Step-by-step explanation:

Given: r = a²t    ...........................(1)

We know the equation,

dp = rdx + sdy

dq = sdx + tdy

Solve these equation we get:

$r = \frac{dp - sdy}{dx} \\t = \frac{dq - sdx}{dy}$

Substitute the value of r and t in equation (1) we get

$\frac{dp - s dy}{dx} = a^{2} \frac{dq - s dx}{dy}$      .............................(2)

According to Monge's Equation:

dp dy - a² dq dx = 0      ......................(3)

dy² - a² dx²  = 0    ..............................(4)

solve equation (3) and (4) by a²-b² we get,

dy - a dx = 0      .................................(5)

dy + a dx = 0      ................................(6)

Solve equation (3) and (5) we get:

dp (a dx) - a² dq dx = 0

dp (a dx) - (a dx) a dq = 0

a dx ( dp - a dq) = 0

dp - a dq = 0     .............................(7)

Integrate equation (5) and (7) we get:

y - dx = A       and     p - a q = B

p - aq = f₁ (y-ax)      ............................(8)

Similarly solve equation (3) and (6) we get:

y + aq = A    and      p + aq = B

p + aq = f₂ (y + ax)     ..........................(9)

Solve equation (8) and (9) we get :

p = $\frac{1}{2}$ [( f₁ (y-ax) + f₂ (y + ax)]

q = $\frac{1}{2a}$ [f₂ (y + ax) -  f₁ (y-ax) ]

Substitute the value of p and q in

dz = pdx + qdy

dz = $\frac{1}{2}$ [f₁ (y-ax) +  f₂ (y + ax)] dx + ($\frac{1}{2a}$) [ f₂ (y + ax) - f₁ (y-ax)]

dz = [$\frac{1}{2a}$ (dy + adx) f₂ (y + ax)] -[ $\frac{1}{2a}$ (dy - adx) f₁ (y-ax)]

Integrate the above equation with respect to dz we get,

z  = f₂ (y + ax) + f₁ (y-ax)

r = a²t  was solved by using Monge's theorem.

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