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Help with Parametric Question

Could you please help me with part ii? I'm not sure what I'm supposed to do. Thank you in advance!

https://scontent-lht6-1.xx.fbcdn.net/v/t1.15752-9/79177865_1065557560450998_5681191759133016064_n.jpg?_nc_cat=107&_nc_ohc=ogHcjPVVejUAQnEhxM3PBf3pTzmoGduV3r1WFG0OgQPA8ZiiHc0rcS09w&_nc_ht=scontent-lht6-1.xx&oh=314a0114cfcc99dd243c1cb97a0f63fc&oe=5E6E8703
Original post by Robertisdank


the gradient of the tangent is equal to tan Θ
Original post by Robertisdank


Find the gradient of the line QP first. You know the coords of Q and also P, so just find the gradient between two points ?
Original post by RDKGames
Find the gradient of the line QP first. You know the coords of Q and also P, so just find the gradient between two points ?

Alright, I've got the gradient ( or tan theta) to be equal to 2t/ t^2 -1 . I'm not sure about the angle part. My intitial thoughts were to somehow use the fact that PR is parallel to the x- axis or to somehow use SOHCAHTOA to manipulate my tan theta into a sin theta or cos theta, but I don't know the exact lengths of the sides of the triangle.
Original post by Robertisdank
Alright, I've got the gradient ( or tan theta) to be equal to 2t/ t^2 -1 . I'm not sure about the angle part. My intitial thoughts were to somehow use the fact that PR is parallel to the x- axis or to somehow use SOHCAHTOA to manipulate my tan theta into a sin theta or cos theta, but I don't know the exact lengths of the sides of the triangle.


Be careful, tan theta is 1/t.

What you have found is gradient PQ which is also equivalent to tan phi.

The simplest thing to do here is just sub in t=1/tan theta into this gradient PQ

Hence, you will obtain tan phi in terms of tan theta, and you should be able to spot a double angle expression sitting there hence being able to show that

tan phi = tan 2theta and and so phi = 2theta.
Original post by RDKGames
Be careful, tan theta is 1/t.

What you have found is gradient PQ which is also equivalent to tan phi.

The simplest thing to do here is just sub in t=1/tan theta into this gradient PQ

Hence, you will obtain tan phi in terms of tan theta, and you should be able to spot a double angle expression sitting there hence being able to show that

tan phi = tan 2theta and and so phi = 2theta.

Ah, okay. I see it now. Thanks for the help!

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