Question #224013

The tangent at the point P(θ) on the ellipse x2/4+y2/3 = 1 passes through the point A(2, 1). Show that √3cosθ + sin θ = √3. Find all the solutions of this equation which are in the range 0 ≤ θ < 2π. Hence find the coordinates of the points of contact of the tangents to the ellipse from A. 


1
Expert's answer
2021-08-16T16:48:02-0400
x24+y23=1\dfrac{x^2}{4}+\dfrac{y^2}{3}=1

x=2cosθ,y=3sinθx=2\cos \theta, y=\sqrt{3}\sin \theta

yx=yθxθ=3cosθ2sinθy_{x}'=\dfrac{y'_{\theta}}{x'_{\theta}}=-\dfrac{\sqrt{3}\cos \theta}{2\sin \theta}

slope=3cosθ2sinθslope=-\dfrac{\sqrt{3}\cos \theta}{2\sin \theta}

y3sinθ=3cosθ2sinθ(x2cosθ)y-\sqrt{3}\sin \theta=-\dfrac{\sqrt{3}\cos \theta}{2\sin \theta}(x-2\cos \theta)

2sinθ(y3sinθ)=3cosθ(x2cosθ)2\sin \theta( y-\sqrt{3}\sin \theta)=-\sqrt{3}\cos \theta(x-2\cos \theta)

The equation of the tangent line at the point P(θ)P(\theta) to the ellipse is


x(3cosθ)+y(2sinθ)=23x(\sqrt{3}\cos \theta)+y(2\sin \theta)=2\sqrt{3}

Point A(2,1)A(2, 1)


2(3cosθ)+1(2sinθ)=232(\sqrt{3}\cos \theta)+1(2\sin \theta)=2\sqrt{3}

3cosθ+sinθ=3\sqrt{3}\cos \theta+\sin \theta=\sqrt{3}

Then


32cosθ+12sinθ=32\dfrac{\sqrt{3}}{2}\cos \theta+\dfrac{1}{2}\sin \theta=\dfrac{\sqrt{3}}{2}

sinπ3cosθ+cosπ3sinθ=32,0θ2π\sin\dfrac{\pi}{3}\cos \theta+\cos\dfrac{\pi}{3}\sin \theta=\dfrac{\sqrt{3}}{2}, 0\leq\theta\leq2\pi

sin(π3+θ)=32\sin(\dfrac{\pi}{3}+\theta)=\dfrac{\sqrt{3}}{2}

π3π3+θ7π3\dfrac{\pi}{3}\leq\dfrac{\pi}{3}+\theta\leq\dfrac{7\pi}{3}

π3+θ=π3=>θ=0\dfrac{\pi}{3}+\theta=\dfrac{\pi}{3}=>\theta=0

Or


π3+θ=2π3=>θ=π3\dfrac{\pi}{3}+\theta=\dfrac{2\pi}{3}=>\theta=\dfrac{\pi}{3}

Or


π3+θ=7π3=>θ=2π\dfrac{\pi}{3}+\theta=\dfrac{7\pi}{3}=>\theta=2\pi

{0,π3,2π}\{0, \dfrac{\pi}{3}, 2\pi\}



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