conics (1 Viewer)

[nrumble42]

so i have two tangents, and I have to find the point of intersection but i cant get anywhere when i do it

tangents:
y(3cosx)-x(2sinx)=6 and y(3sinx) +x(2cosx)=6

 

[InteGrand]

so i have two tangents, and I have to find the point of intersection but i cant get anywhere when i do it

tangents:
y(3cosx)-x(2sinx)=6 and y(3sinx) +x(2cosx)=6

Are you sure those should be cos(x) and sin(x) instead of something like cos(theta) and sin(theta)? They wouldn't be straight lines as you've written them.

 

[nrumble42]

Are you sure those should be cos(x) and sin(x) instead of something like cos(theta) and sin(theta)? They wouldn't be straight lines as you've written them.

ahh yeah sorry, it does say theta

 

[nrumble42]

bump

 

[InteGrand]

bump

Now that those are cos(theta) etc., all you need to do is find the intersection of two lines. This can be done by solving for x in one of the lines (in terms of y), and plugging this into the other line's equation (i.e. standard procedure of solving two simultaneous linear equations in two unknowns).

 

[pikachu975]

so i have two tangents, and I have to find the point of intersection but i cant get anywhere when i do it

tangents:
y(3cosx)-x(2sinx)=6 and y(3sinx) +x(2cosx)=6

Rearranging first tangent equation for x in terms of y:
3ycosA - 2xsinA = 6
2xsinA = 3ycosA - 6
x = 3(ycosA - 2)/2sinA

Sub this into second equation
3ysinA + 2xcosA = 6
3ysinA + 3cosA(ycosA-2)/sinA = 6
3ysin^2 A + 3ycos^2 A - 6cosA = 6sinA
3y (sin^2 A + cos^2 A) = 6(sinA + cosA)
y = 2(sinA+cosA)

Sub y into the first tangent equation
6cosA(sinA+cosA) - 2xsinA = 6
6sinAcosA + 6cos^2 A - 2xsinA = 6
6sinAcosA + 6 - 6sin^2 A - 2xsinA = 6
6sinAcosA - 6sin^2 A = 2xsinA
3cosA - 3sinA = x (sinA doesn't = 0)
x = 3(cosA-sinA)

Therefore they intersect at x = 3(cosA-sinA), y = 2(sinA+cosA)