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1. Let C a parabola of focus F and directrix L. Let N be a unit normal vector to L and assume that C contains a point X ∈ V

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L. A. G. test n. 3, april 21, 2016 Name:

1. Let C a parabola of focus F and directrix L. Let N be a unit normal vector to L and assume that C contains a point X ∈ V

2

such that k X − F k= 4 and the cosine of the angle between X − F and N is equal to −

15

.

(a) Assuming that F is in the negative half-plane determined by N , compute d(F, L).

(b) Assuming that F is in the positive half-plane determined by N , compute d(F, L).

Solution. (a) In this case d(F, L) +

45

= 4. Hence d(F, L) =

165

(you can also check this with the polar equation).

(b) In this case 4 +

45

= d(F, L). Hence d(F, L) =

245

.

2. (a) Let P

1

= (−1, 1, −1), P

2

= (1, 2, 2), P

3

= (−3, 0, −4), P

4

= (1, 1, 1). Is there a plane containing these four points? If yes, write down its cartesian equation, if no explain why.

(b) Let P

1

= (−1, 1, −1), P

2

= (1, 2, 2), Q

3

= (1, 1, 2), Q

4

= (1, 1, 1). Is there a plane containing these four points? If yes, write down its cartesian equation, if no explain why.

Solution. (a) Yes, x + y − z = 1 (note: the first three points are collinear). (b) No: the

unique plane containing the first three points has equation 3x − 2z = −1, and the fourth

point is not contained in it.

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