FP1 question

the normal to the hyperbola xy = c^2 at P(ct,c/t) has eqn y = (t^2)x + c/t - ct^3

the normal to the hyperbola at P meets the hyperbola again at the point Q.

find in terms of t, the coordinates of the point Q.

the answer is (-c/t^3, -ct^3).

I am completely lost as to how to solve this.

Solve the equation of the hyperbola simultaneously with the equation of the normal. This will give you Q (and P).

Reply 2

Lunch_Box

OP

15

Original post by Mr M

Solve the equation of the hyperbola simultaneously with the equation of the normal. This will give you Q (and P).

Is there an easier way to solve this without having to use t1 and t2?

... (ct₂) t₁³ - (c/t₂) t₁ - c t₁⁴ + c = 0

∴ t₂ t₁³ - (t₁ / t₂) - t₁⁴ + 1 = 0

∴ ( t₂ t₁³ + 1 ) - (t₁ / t₂)( 1 + t₂ t₁³ ) = 0

∴ ( t₂ t₁³ + 1 ) ( 1 - (t₁ / t₂)) = 0, ... t₁ ≠ t₂

∴ t₂ t₁³ + 1 = 0

∴ t₂ = -1 / t₁³

∴ normal to xy = c² at P( ct₁, c/t₁ )

meets the curve again in P' ( -c/ t₁³ , -c t₁³ ).

+ I can't see how this will come up in an FP1 exam. Not mentioned in textbook or by teacher.

Reply 3

mathsman77

7

Original post by Lunch_Box

Is there an easier way to solve this without having to use t1 and t2?

... (ct₂) t₁³ - (c/t₂) t₁ - c t₁⁴ + c = 0

∴ t₂ t₁³ - (t₁ / t₂) - t₁⁴ + 1 = 0

∴ ( t₂ t₁³ + 1 ) - (t₁ / t₂)( 1 + t₂ t₁³ ) = 0

∴ ( t₂ t₁³ + 1 ) ( 1 - (t₁ / t₂)) = 0, ... t₁ ≠ t₂

∴ t₂ t₁³ + 1 = 0

∴ t₂ = -1 / t₁³

∴ normal to xy = c² at P( ct₁, c/t₁ )

meets the curve again in P' ( -c/ t₁³ , -c t₁³ ).

+ I can't see how this will come up in an FP1 exam. Not mentioned in textbook or by teacher.

It isn't on the OCR syllabus for FP1 (not that I know of anyway), but it isn't as difficult as it looks, if you rearrange xy=c² and sub into the equation of the normal, collect like terms you get coeffcient of y to be
ct³ - c/t and the term on the end to be c²t²

The question tells you one of the solutions to the quadratic (y - c/t) and you use this work out what the other solution is. (if you've done it correctly, it is the part of the coeff of y that isn't - c/t :tongue:

Reply 4

insparato

15

I did this question may about 7 years ago, haven't attempted it yet, so forgive me if this doesn't help you just throwing my 2 cents in.

You may find

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

useful, when dotting up the algebra. Coordinates system questions I found were notorious for the algebra you had to crunch.

I may help you with other coordinate systems questions involving cubics anyways!

FP1 question

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