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1.5.16 Gradient of a parametric curve (A-level only)

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Question 27

The path of a beam of light reflecting off a parabolic mirror is modeled by the parametric equations

x=2.5t2 and y=5t,−4≤t≤4 x = 2.5t^2 \text{ and } y = 5t, \quad -4 \le t \le 4 x=2.5t2 and y=5t,−4≤t≤4

where x x\,x and y y\,y are measured in centimeters.

a.

Determine the Cartesian equation of the reflective surface in the form y2=f(x)y^2 = f(x)y2=f(x).

[2]
bi.

A specific photon strikes the mirror at point AAA where the parameter t=at = at=a (with a>0,a≠1a > 0, a \neq 1a>0,a=1). The tangent to the curve at point AAA makes an angle θ\thetaθ with a line through AAA parallel to the xxx-axis. The point BBB is located at (2.5,0)(2.5, 0)(2.5,0). The line segment ABABAB makes an angle ϕ\phiϕ with the positive xxx-axis.

By calculating the gradient of the curve, show that tan⁡θ=1a\tan \theta = \frac{1}{a}tanθ=a1​.

[3]
bii.

Find an expression for tan⁡ϕ\tan \phitanϕ in terms of aaa, simplifying your answer.

[3]
biii.

Hence, show that tan⁡2θ=tan⁡ϕ\tan 2\theta = \tan \phitan2θ=tanϕ.

[3]

1.5.16 Gradient of a parametric curve (A-level only) Questions

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