I am studying maths as a hobby and am struggling with this question:
Show that the equation of the tangent to the curve $x=4\cos\theta,y=2\sin\theta$ at the point with parameter $\theta$ is $x\cos\theta+2y\sin\theta=4$. Hence find the equations of the tangents which pass through the point $(5,0)$ and the coordinates of their points of contact.
I have said:
equation of curve: $x=4\cos\theta, y=2\sin\theta$
$\frac{dx}{d\theta} = -4\sin\theta, \frac{dy}{d\theta} = 2cos\theta$
$\frac{dy}{dx}=\frac{dy}{d\theta}.\frac{d\theta}{dx}=\frac{dy}{d\theta}\div \frac{dx}{d\theta} = \frac{2\cos\theta}{-4\sin\theta}$
$\rightarrow \frac{y-2\sin\theta}{x-4\cos\theta}=\frac{2\cos\theta}{-4sin\theta}$
$\rightarrow x\cos\theta +2\sin\theta y=4$
I have now reasoned as follows:
The gradient of the line which passes through (5,0) is $\frac{y-0}{x-5}$
To find the gradient of the tangent I need to differentiate it:
$\rightarrow x\cos\theta +2\sin\theta y=4$
$-\sin x +2\sin\theta.\frac{dy}{dx}+2\cos\theta y=0$
$\rightarrow \frac{dy}{dx}=\frac{\sin x-2\cos\theta y}{2 \sin \theta}$
$\rightarrow \frac{y}{x-5} = \frac{\sin x-2\cos\theta y}{2 \sin \theta}$
But I cannot see how to proceed from here, and feel sure I am going wrong.
I don't think you need to differentiate: You have that the general tangent line to the curve has equation $$x\cos\theta+2y\sin\theta=4$$ and you need $(5,0)$ to belong to the tangent, so it has to be $$5\cos\theta=4\rightarrow \cos\theta=\frac{4}{5}$$ and using $\sin\theta=\pm\sqrt{1-\cos^2\theta}=\pm\frac{3}{5}$, so the equation is $$4x\pm 6y=20$$ the sign depends on the value of $\theta$ (the point $(5,0)$ belongs to the tangent line at the point $P(\theta)=(x(\theta),y(\theta))$ and $P(-\theta)$, where $P$ is a point in the curve as a function of the parameter $\theta$)