Parametric Curves

[heady night]

I have managed to find that dy/dx=-sin(theta)=-x/4 and the Cartesian equation of the curve is y=1-(x^2)/8 but I don't know where to go from there.

[sinful drum]

That's correct.
Let's write the equation of tangent to y = 1 - x^2/8 at x = a.
f(x) = 1 - x^2/8
f'(x) = -x/4
So, the tangent is:
y = f'(a)(x - a) + f(a)
y = -(a/4)(x - a) + 1 - a^2/8
y = -(a/4)x + (a^2/8 + 1)
We know that this line must pass through (3, 0). So, subtitute those coordinates to find a.

Remember to check whether the resulting values lie on the curve.

[heady night]

so the equation of the tangent is y=-(3/4)x+9/4?

[sinful drum]

No.

[heady night]

y-y1=m(x-x1) so y-0=(-x1/4)(x-3)?

where x1=3

[sinful drum]

Well, as I said above, the equation of tangent is y = -(a/4)x + (a^2/8 + 1).
It must pass through (3, 0), so we must have:
0 = -3(a/4) + (a^2/8 + 1)
a^2/8 - 3a/4 + 1 = 0
Solve this quadratic equation to find the values of a.

[heady night]

so a is the x coord of P?

[sinful drum]

Yes.

[heady night]

what does it mean when there is two possible values of a? is it that P could be one of two places?

[sinful drum]

In general, yes.
However, note that your function is only defined for 0 ≤ x < 4.

[heady night]

but if it was defined for all R, P could be in two places such that the tangent at P could pass through (3,0)?

[sinful drum]

Yes.

[heady night]

I get it now thanks a lot😄

[sinful drum]

You're welcome!

[heady night]

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