Question

using lagrange's multiplier method, find the shortest distance between the line y=10−2x and the ellipse x^2/4+y^2/9=1.​

Answer 1

Step-by-step explanation:

Let the point on ellipse be (2cosθ,3sinθ)(2cos⁡θ,3sin⁡θ)

Let F=(x−2cosθ)2+(y−3sinθ)2+α(2x+y−10)F=(x−2cos⁡θ)2+(y−3sin⁡θ)2+α(2x+y−10).

I partially diffentiated FF wrt xx,yy and θθ equated to 00.

I get tanθ=34tan⁡θ=34 and α=−2α=−2

The point on ellipse I get is (85,95)(85,95) and the point on line I get is (185,145)(185,145)

Is this method correct? Can I take θθ to be an

Answer 2

Answer:

ur answers is here

Step-by-step explanation:

Let the point on ellipse be (2cos0,3sine) (2cos0,3sine)

Let F=(x-2cos0)2+

(y-3sin0)2+a(2x+y-10)F=(x-2cos0)2+ (y-3sin0)2+a(2x+y-10).

I partially

diffentiated FF wrt xx,yy and 00 equated to 00.

I get tane-34tan0-34 and a=-2a=-2

The point on ellipse I get is (85,95) (85,95) and the point on line I get is (185,145) (185,145)

Is this method correct? Can I take 00 to be

an