I'm having difficulties with the question below. It is from a Cambridge past paper.
The equation of a curve is $xy=12$ and the equation of a line $l$ is $2x+y=k$, where $k$ is a constant. In the case where $k=10$, one of the points of intersection is $P(2,6)$. Find the angle, in degrees correct to $1$ decimal place, between $l$ and the tangent to the curve at $P$."
I know the answer is $8.1$ degrees and the method to find the answer is either by using differentiation or by using the fact that $\text{gradient}=\tan(\theta)$.
Thank you
You will need to use both differentiation and the fact that the gradient$=\tan(\theta)$. For the curve you have:$$xy=12\implies y=12x^{-1}$$Differentiate this to find the gradient of the tangent line at $x=2$. The line is given by:$$2x+y=10$$So you should know what its gradient is equal to.
Lets call the angle that the line makes with the x-axes $\alpha$ and the angle that the tangent to the curve makes with the x-axes $\beta$. We therefore know the values of $\tan(\alpha)$ and $\tan(\beta)$ since these are equal to the gradients of the line and the tangent to the curve at $x=2$.
You now just need to use the angle difference formula:$$\tan(\alpha-\beta)=\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$$to work out the tan of the angle between these two. Then find the arctan of this value to finally give you the angle between them.