# The tangent and the normal to the conic x^2/a^2+y^2/b^2=1 at a point (acostheta, bsintheta) meet the major axis in the points P and P', where PP'=a Show that e^2cos^2theta + costheta -1 = 0, where e is the eccentricity of the conic?

## The tangent and the normal to the conic ${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$ at a point $\left(a \cos \theta , b \sin \theta\right)$ meet the major axis in the points $P$ and $P '$, where $P P ' = a$ Show that ${e}^{2} {\cos}^{2} \theta + \cos \theta - 1 = 0$, where $e$ is the eccentricity of the conic

Mar 24, 2017

See below.

#### Explanation:

At point ${x}_{0} , {y}_{0}$ there is a tangent line and a normal line given by

${L}_{t} \to y = {y}_{0} + {m}_{0} \left(x - {x}_{0}\right)$ with

${x}_{0} = a \cos {\theta}_{0}$
${y}_{0} = b \sin {\theta}_{0}$

${m}_{0} = {\left(\frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}}\right)}_{{\theta}_{0}} = - \frac{b}{a} \frac{\cos {\theta}_{0}}{\sin {\theta}_{0}}$

and

${L}_{n} \to y = {y}_{0} - \frac{1}{m} _ 0 \left(x - {x}_{0}\right)$

Their intersections with the $x$ axis are given by

${p}_{t} = \left(\frac{{m}_{0} {x}_{0} - {y}_{0}}{m} _ 0 , 0\right)$
${p}_{n} = \left({x}_{0} + {m}_{0} {y}_{0} , 0\right)$

and their distance

$\left\lVert {p}_{n} - {p}_{t} \right\rVert = \frac{{b}^{2} \cos {\theta}_{0} + {a}^{2} \sin {\theta}_{0} \tan {\theta}_{0}}{a}$

but $\left\lVert {p}_{n} - {p}_{t} \right\rVert = a$ so

${b}^{2} \cos {\theta}_{0} + {a}^{2} \sin {\theta}_{0} \tan {\theta}_{0} = {a}^{2}$

or

${b}^{2} {\cos}^{2} {\theta}_{0} + {a}^{2} {\sin}^{2} {\theta}_{0} = {a}^{2} \cos {\theta}_{0}$

or

$\left({b}^{2} / {a}^{2} - 1\right) {\cos}^{2} {\theta}_{0} - \cos {\theta}_{0} + 1 = 0$

or

${e}^{2} {\cos}^{2} {\theta}_{0} - \cos {\theta}_{0} + 1 = 0$