Conics question and does anyone have sk patel worked solutions? (1 Viewer)

[mysterymarkplz]

The question is,
3)V and V' are the feet of the perpendiculars from S and S' respectively to the tangent at P(X1, Y1) to the ellipse 4x^2 + 9y^2 = 36.
Prove that:
a) SV.S'V' = 4
b)V and V' lie on the auxiliary circle x^2 + y^2 = 9

I used a really dodgy method of doing this question, I put the tangent at the point 0,2 and then get the coordinates of V and V' then i solved a and b from there, am i allowed to do this in the test i mean i'm not assuming anything i did put it on a known point on the ellipse but my method seems really weird. Is there any other way to do this question?

 

[Carrotsticks]

For (a):

1. Express tangent in general form.

2. Find coordinates of foci.

3. Use perpendicular distance formula twice, one to find SV and another to find S'V'.

4. Multiply the two distances, a lot will cancel out.

For (b):

Consider the following diagram.

1. Prove that triangle PVS is congruent to triangle PVT (thus SV = VT)

2. Prove that triangle OSN is similar to triangle S'TS (use SAS since SV/ST is already in ratio 1/2 and so is SO/SS')

3. Prove that S'T = 6 (using PS + PS' = 2a, but PS = PT so PT + PS' = 2a)

4. By similar triangle ratios, OV/S'T = 1/2, but S'T = 6, so OV = 3 (hence lies on circle x^2+y^2=9)

5. Similarly, for OV' = 3, hence V and V' lie on the circle x^2+y^2=9.