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Quick Question Ellipses (1 Viewer)

cutemouse

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Hello,

I have a quick question... I was doing a question on ellipses in an example and it says that

All lines perpendicular to 6x+2y+3=0 are of form 2x-6y=k....

I don't get why this is and when I rearrange it I absolutely fudge it. Could someone please tell me a quick way to work this out without having to go through the whole rearranging process that I somehow always screw up?

Thanks

EDIT: The whole question I am referring to:

Determine the equations of the tangents to the ellipse x2+16y2=25 which are perpendicular to the line 6x+2y+3=0, and find their points of contact with the curve
 
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lychnobity

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The gradient of a line in general form is m= -a/b (ie line is ax + by + c = 0)

Therefore, if you want to get the line perpendicular to it, put the value of b with the x and the a with y.
 

Drongoski

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This is a 2U question.

gradient m1 of 6x + 2y + 3 = 0 is -3

gradient m2 of 2x - 6y = k is 1/3

m1 x m2 = -1

so the 2 lines are perpendicular

by varying k you generate the whole set of lines perp to line-1
 

cutemouse

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This is a 2U question.
Stop trying to signature flash. It's also a part of a 4U question, as you would know if you had read my whole post, and you didn't answer my question. I was asking HOW I can obtain the equation (ie. how to find lines perp. to a given line)
 
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cutemouse

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The gradient of a line in general form is m= -a/b (ie line is ax + by + c = 0)

Therefore, if you want to get the line perpendicular to it, put the value of b with the x and the a with y.
Hi,

Thanks for your reply... But I don't get why the 3 on the one side becomes a 'k' on the other side, ie. shouldn't it be (-k)? If not why?
 

cutemouse

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Hi again,

Wouldn't x-3y=k be also perp. to 6x+2y+3=0??? So why is keeping the constant also important?

Here's the whole question:

Determine the equations of the tangents to the ellipse x2+16y2=25 which are perpendicular to the line 6x+2y+3=0, and find their points of contact with the curve

The solution rearranges the line perp to 6x+2y+3=0 to y=(2x-k)/6, which is where my question above arises from.

Thanks in advanced
 

Drongoski

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Stop trying to signature flash. It's also a part of a 4U question, as you would know if you had read my whole post, and you didn't answer my question. I was asking HOW I can obtain the equation (ie. how to find lines perp. to a given line)
jm01

Sorry. I don't understand yr question then. Just disregard what I posted.
 

GUSSSSSSSSSSSSS

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k, -k all the same, its just a constant

so basically u just find the gradient of the perpendicular line, which drongoski was rite in saying is a 2u thing

then just put the derivative of the ellipse equal to this

..
 

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