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Evening all. Just finished Ex 4c but for two questions... Any help greatly appreciated as always.

17.

The normal at the point P(2t,2/t) to the curve with equation xy=4 meets the lines with equations y=x and y=-x at the points Q and R repectively. Prove that PQ=PR.

I managed to get it down to two very similiar, but not equal, lengths in t....

20.

The tangents at P(ct1,c/t1) and Q(ct2,c/t2) to the rectangular hyperbola with equation xy=c^2 meet on the rectangular hyperbola with equation xy=c^2/4. Prove that PQ is a tangent to the curve with equation xy=4c^2.

I'm not sure how the seperate parts of the questions relate.

Thanks!

17.

The normal at the point P(2t,2/t) to the curve with equation xy=4 meets the lines with equations y=x and y=-x at the points Q and R repectively. Prove that PQ=PR.

I managed to get it down to two very similiar, but not equal, lengths in t....

20.

The tangents at P(ct1,c/t1) and Q(ct2,c/t2) to the rectangular hyperbola with equation xy=c^2 meet on the rectangular hyperbola with equation xy=c^2/4. Prove that PQ is a tangent to the curve with equation xy=4c^2.

I'm not sure how the seperate parts of the questions relate.

Thanks!

(17)

On the curve, y + x dy/dx = 0. So the tangent at P has gradient -(2/t)/(2t) = -1/t^2. The normal at P has gradient t^2 and equation y = t^2 (x - 2t) + 2/t.

At Q,

t^2 (x - 2t) + 2/t = x

x (t^2 - 1) = 2t^3 - 2/t = (2/t)(t^4 - 1) = (2/t)(t^2 - 1)(t^2 + 1)

x = (2/t)(t^2 + 1) = 2t + 2/t

y = 2t + 2/t

So

|PQ|^2

= (2t + 2/t - 2t)^2 + (2t + 2/t - 2/t)^2

= 4/t^2 + 4t^2

At R,

t^2 (x - 2t) + 2/t = -x

x (t^2 + 1) = 2t^3 - 2/t = (2/t)(t^4 - 1) = (2/t)(t^2 - 1)(t^2 + 1)

x = (2/t)(t^2 - 1) = 2t - 2/t

y = -2t + 2/t

So

|PR|^2

= (2t - 2/t - 2t)^2 + (-2t + 2/t - 2/t)^2

= 4/t^2 + 4t^2

On the curve, y + x dy/dx = 0. So the tangent at P has gradient -(2/t)/(2t) = -1/t^2. The normal at P has gradient t^2 and equation y = t^2 (x - 2t) + 2/t.

At Q,

t^2 (x - 2t) + 2/t = x

x (t^2 - 1) = 2t^3 - 2/t = (2/t)(t^4 - 1) = (2/t)(t^2 - 1)(t^2 + 1)

x = (2/t)(t^2 + 1) = 2t + 2/t

y = 2t + 2/t

So

|PQ|^2

= (2t + 2/t - 2t)^2 + (2t + 2/t - 2/t)^2

= 4/t^2 + 4t^2

At R,

t^2 (x - 2t) + 2/t = -x

x (t^2 + 1) = 2t^3 - 2/t = (2/t)(t^4 - 1) = (2/t)(t^2 - 1)(t^2 + 1)

x = (2/t)(t^2 - 1) = 2t - 2/t

y = -2t + 2/t

So

|PR|^2

= (2t - 2/t - 2t)^2 + (-2t + 2/t - 2/t)^2

= 4/t^2 + 4t^2

deity47

...

20.

The tangents at P(ct1,c/t1) and Q(ct2,c/t2) to the rectangular hyperbola with equation xy=c^2 meet on the rectangular hyperbola with equation xy=c^2/4. Prove that PQ is a tangent to the curve with equation xy=4c^2.

I'm not sure how the seperate parts of the questions relate.

Thanks!

20.

The tangents at P(ct1,c/t1) and Q(ct2,c/t2) to the rectangular hyperbola with equation xy=c^2 meet on the rectangular hyperbola with equation xy=c^2/4. Prove that PQ is a tangent to the curve with equation xy=4c^2.

I'm not sure how the seperate parts of the questions relate.

Thanks!

You use the first part about the tangents meeting to define a relationship beytween t1 and t2.

THEN you do the other bit. Get eqn of the line PQ and interesect it with the other hyperbola.

Jonny W

2t^3 - 2/t = (2/t)(t^4 - 1)

2t^3 - 2/t = (2/t)(t^4 - 1)

Fermat

You use the first part about the tangents meeting to define a relationship beytween t1 and t2.

THEN you do the other bit. Get eqn of the line PQ and interesect it with the other hyperbola.

THEN you do the other bit. Get eqn of the line PQ and interesect it with the other hyperbola.

Thanks again!

- Edexcel AS Statistics and Mechanics 2024
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- biology past paper question 2019 june/may paper 1
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