Conics questions (1 Viewer)

[sonnylongbottom]

1.Prove that the line 3x+4y=10 is a normal to the hyperbola 2x^2-3y^2=5 and find the point at which the line is normal.

2. P is any point on the ellipse x^2/a^2 + y^2/b^2 = 1 with foci S and S', prove that PS + PS' = 2a

i suck at conics so yeah help please!

 

[midifile]

2. P is any point on the ellipse x^2/a^2 + y^2/b^2 = 1 with foci S and S', prove that PS + PS' = 2a

Okay. This is gonna be a bit harder without a diagram but here goes.

Let the points M and M' be points on the directricies of the ellipse with the same y value as P. So if P = (x, y), M = (a/e, y) and M'= (a/e,y)
Now,
In an ellipse, PS/PM = e
so PS = ePM and PS' = ePM'
therefore PS + PS' = e(PM + PM')
If you look on your diagram you will see that this is a straight line from -a/e to a/e
So PS + PS'= e x (2a/e)
= 2a

Im not sure about the first one. It would be easy if it was a tangent because then you could prove that they only touch once but its a normal so that wont work

 

[3unitz]

1.Prove that the line 3x+4y=10 is a normal to the hyperbola 2x^2-3y^2=5 and find the point at which the line is normal.

d/dx (2x^2-3y^2) = d/dx 5
4x - d/dy (3y^2) dy/dx = 0
4x - 6y dy/dx = 0
dy/dx = 4x/6y

3x+4y=10 => m = - 3/4

.'. for normal, gradient of the tangent at the point on the hyperbola has to be equal to 4/3. to find this point on the hyperbola we solve 4/3 = 4x/6y, and 2x^2-3y^2=5 simultaneously.

4/3 = 4x/6y
2y = x

2x^2-3y^2=5
2(2y)^2 - 3y^2 = 5
y^2 = 5
y = sqrt 5
x = 2 sqrt 5

EDIT: just point (2sqrt 5, sqrt 5)

(-2sqrt 5, -sqrt 5) doesnt lie on the line 3x+4y=10

 

[minijumbuk]

3unitz, didn't you only do 3 units? xD