Conics- Ellipse Question (1 Viewer)

[hunter1]

Hi, i am having a few problem with the final parts of this question.

Any help is greatly appreciated

 

[ssglain]

For part (v), use the gradients of tangents at P & Q and the formula:
tan(theta) = |(m<SUB>1 </SUB>-<SUB> </SUB>m<SUB>2</SUB>)/(1 - m<SUB>1</SUB>m<SUB>2</SUB>)|
Work through the algebra and you will get tan(theta) = 12/[5sin(2theta)]

Keep in mind that b<SUP>2 </SUP>= a<SUP>2</SUP>(1 - e<SUP>2</SUP>) => e<SUP>2</SUP> = 1 - b<SUP>2</SUP>/a<SUP>2</SUP>
i.e. 4 = 9(1 - e<SUP>2</SUP>) -> 2/3 = SQRT(1 - e<SUP>2</SUP>)
and e<SUP>2</SUP> = 1 - 4/9 -> e<SUP>2</SUP> = 5/9

Now if you divide both the numerator and denominator of 12/[5sin(2theta)] by 9, the top constant becomes 4/3, which is 2SQRT(1 - e<SUP>2</SUP>) and the bottom constant becomes 5/9, which is e<SUP>2</SUP>.

Therefore, tan(theta) = [2SQRT(1 - e<SUP>2</SUP>)]/[e<SUP>2</SUP>sin(2theta)] as required.

 

[hunter1]

Thanks

 

[pLuvia]

i) Q has the parameter @+pi/2, so using the general parametric equation of an ellipse
Q(3cos(@+pi/2), 2sin(@+pi/2))
=Q(-3sin@, 2cos@)

ii) Just use distance formula after finding out was P is

Spoiler

P(3cos@, 2sin@)

iii) You can actually derive the general tangent formula for an ellipse but using implicit differentiation or normal differentiation. After you have done this, sub in the points P and Q, now you get two equations. Solve them simultaneously and you get the coordinates of T

iv) After you get the coordinates of T, you get something in parametric form. Split the coordinates so that you have x= blah and y=bleh.

Square both of these then add them both together. i.e. x²+y²=blah²+bleh²

Simplify this a bit and you should automatically see this is an ellipse

Sorry don't have that much time to actually type up the working, have uni exams soon you see. If you are really stuck I'll put it up, but try it first

Good luck