Find the coordinates of the point where the normal cuts the curve again problem

Question

Find the equation of the normal to the curve $x=2\cos\theta$, $y=3\sin\theta$ at the point where $\theta=\frac{1}{4}\pi$. Find the coordinates of the point where this normal cuts the curve again.

Okay so I found the equation of the normal to the curve by using $y-y_1=m(x-x_1)$ and got:

$6y=4x+5\sqrt{2}$.

It's the bit of the question in bold that I'm confused about.

I know that all points for which $x=2\cos\theta$ and $y=3\sin\theta$ are on the curve. So any point on the normal and on the curve will satisfy these two equations and the equation of the normal I found.

So subbing in $x=2\cos\theta$ and $y=3\sin\theta$ into the equation of the normal I get:

$18\sin\theta=8\cos\theta+5\sqrt{2}$

I'm a bit puzzled as to how to use this bit of information to find the coordinates of the point where the normal cuts the curve again. I can see some potential solutions of $\theta$ for the above equation such as $\frac{9\pi}{4}$ which would give coordinates of $(\sqrt{2},\frac{3\sqrt{2}}{2})$ but I'm confused as my textbook gives the answer as $(-\frac{137}{97}\sqrt{2},-\frac{21}{194}\sqrt{2})$

Help/guidance much appreciated.

Answer 1

I think what you've done is correct.

Now, squaring the both sides of $$18\sin\theta=8\cos\theta+5\sqrt 2\tag 1$$ gives $$18^2(1-\cos^2\theta)=(8\cos\theta+5\sqrt 2)^2$$ Solving this for $\cos\theta$ will give you $$\cos\theta=-\frac{137}{97\cdot 2}\sqrt 2,\ \ \frac{1}{\sqrt 2}$$ From here, you can get $\sin\theta$. Then, choose $(\cos\theta,\sin\theta)$ which satisfies $(1)$.

Answer 2

Assuming you've done everything correctly so far, perhaps you could use the sum to product formulae? Rearrange your simultaneous equation to $$ 18\sin \theta - 8\cos \theta = 5\sqrt{2}$$ Expand $R\sin(\theta - \alpha)$ and compare the coefficients of $\sin \theta$ and $\cos \theta$ with the LHS above: $$\begin{aligned} & R\cos\alpha = 18 \\ & R\sin\alpha = 8 \end{aligned} \implies \begin{aligned} & R^2 = 18^2 + 8^2 \\ & \tan \alpha = 4/9 \end{aligned}$$

Answer 3

I don't like trigonometric identities.

The equation of the normal at the given point is indeed $6y = 4x + 5\sqrt{2}$. Solving for $y$ yields $y = {2 \over 3}x + {5\sqrt{2} \over 6}$ (1).

Then, writing the ellipse in Cartesian form, we have ${x^2 \over 4} + {y^2 \over 9} = 1$ (2).

Now all is required is to sub (1) into (2), and use the quadratic formula to solve for $x$-components of the two intersections. Substituting the $x$-component of an intersection into (1) will then yield the corresponding $y$-component.