Answer to Question #224013 in Analytic Geometry for Bless

Question

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. 

Answer 1

$$\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$$

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

$$y_{x}'=\dfrac{y'_{\theta}}{x'_{\theta}}=-\dfrac{\sqrt{3}\cos \theta}{2\sin \theta}$$

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

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

$$2\sin \theta( y-\sqrt{3}\sin \theta)=-\sqrt{3}\cos \theta(x-2\cos \theta)$$

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

$$x(\sqrt{3}\cos \theta)+y(2\sin \theta)=2\sqrt{3}$$

Point $A(2, 1)$

$$2(\sqrt{3}\cos \theta)+1(2\sin \theta)=2\sqrt{3}$$

$$\sqrt{3}\cos \theta+\sin \theta=\sqrt{3}$$

Then

$$\dfrac{\sqrt{3}}{2}\cos \theta+\dfrac{1}{2}\sin \theta=\dfrac{\sqrt{3}}{2}$$

$$\sin\dfrac{\pi}{3}\cos \theta+\cos\dfrac{\pi}{3}\sin \theta=\dfrac{\sqrt{3}}{2}, 0\leq\theta\leq2\pi$$

$$\sin(\dfrac{\pi}{3}+\theta)=\dfrac{\sqrt{3}}{2}$$

$$\dfrac{\pi}{3}\leq\dfrac{\pi}{3}+\theta\leq\dfrac{7\pi}{3}$$

$$\dfrac{\pi}{3}+\theta=\dfrac{\pi}{3}=>\theta=0$$

Or

$$\dfrac{\pi}{3}+\theta=\dfrac{2\pi}{3}=>\theta=\dfrac{\pi}{3}$$

Or

$$\dfrac{\pi}{3}+\theta=\dfrac{7\pi}{3}=>\theta=2\pi$$

$$\{0, \dfrac{\pi}{3}, 2\pi\}$$