This is how I did it
you already had from previous parts
x1=(2h)tan
y1=(2h)cos^2
x1^2=((2h)^2) x tan^2
x1^2=((2h)^2) x sec^2 - 1 using (tan^2+1=sec^2)
x1^2=((2h)^2) x 1/cos^2 - 1
x1^2=((2h)^2) x 1/y1/2h -1
x1^2=((2h)^2) x 2h/y1 -1
squaring both sides then gives
x1= 2h x Square root of 2h/y1...
still prepare an adaptable essay if you're into that shit because there is usually an extended response at the end worth half of the marks
but other than that you have to know the text well, visual tecnhiques and what the module entails, it's a bit tricky to prepare any other way lol