Tangent and normal to a point on the ellipse

Question

I'm working through Stroud's Engineering Mathematics for 'fun' and got stuck on the second part of Q3 in Further Problems 8:

"Obtain the equations of the tangent and normal to the ellipse $\frac{x^2}{169} + \frac{y^2}{25} = 1$ at the point $\left(13\cos θ, 5\sin θ\right)$. If the tangent and normal meet the x-axis at the points T and N respectively, show that ON.OT is constant, O being the origin of coordinates."

So I've got the correct equations for the tangent and normal, $\frac{x\cos θ}{13}+\frac{y\sin θ}{5}=1$ and $5y=13x\tan θ-144\sin θ$

Next I've assumed $y=0$ and $θ=0$ if we're intersecting the x-axis (this is correct right?) giving $x\cos θ=x=13$ for the tangent and $13x\tan θ=144\sin θ$ which drops out to $x=\frac{144}{13}$ for the normal. That gives $ON.OT = 13-\frac{144}{13}=\frac{25}{13}$ The answer is supposed to be $144$ though.

If I visualise a point rotating with $θ$ round an ellipse though, the tangent and normal are not going to have a constant separation though (e.g. at $θ=\frac{π}2$ the tangent never intercepts the x-axis). I think I'm misunderstanding something fundamental here (but whenever I think that it's usually a good indicator I've done something trivially stupid). I've a hunch it has to do with my concepts around parametric forms (never been comfortable with them to be honest).

Any hints or explanations? Don't mind spoilers as I've spent too long on this question already to be honest.

Answer 1

equation of tangent to ellipse(in parametric form)

$bxcos\theta +aysin\theta=ab$ and it cuts x axis at

$x=\dfrac{a}{cos\theta}\tag{1}$

similarly equation of normal to ellipse(in parametric form)

$axsec\theta-bycosec\theta=a^2-b^2$ and it cuts x axis at

$\dfrac{a^2-b^2}{a sec\theta}\tag{2}$

multiply 1 and 2 and you will get

$a^2-b^2=(169-25)=144=constant$

edit:

your equations of tangent and normal both were right but the problem is that, you were assuming $y,\theta $ both equal to $0$ which is wrong and the right way is

put only y=0 and not $\theta =0$ for finding intersection point of tangent and normal (or any line or even any curve in Cartesian form) with x-axis

Do not confuse yourself with this $\theta $ as $\theta$ of polar co-ordinate system where $\theta=0\implies$ horizontal axis

**In case of ellipse the parametric angle $\theta$ is not taken with respect to x-axis (it is called eccentric anomaly)

so you can't put $\theta=0$ for finding intersection with x-axis .**

Answer 2

HINT just put $y=0$ so that you get the coordinates of $N$ and $T$ in terms of $\theta$