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

Question

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

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Answer 1 · 2017-03-24T12:51:20

See below.

At point $x_0,y_0$ there is a tangent line and a normal line given by

$L_t->y=y_0+m_0(x-x_0)$ with

$x_0 = a cos theta_0$
$y_0=b sin theta_0$

$m_0=(((dy)/(d theta))/((dx)/(d theta)))_(theta_0)=-b/a (costheta_0)/(sintheta_0)$

and

$L_n->y = y_0 -1/m_0(x-x_0)$

Their intersections with the $x$ axis are given by

$p_t = ((m_0x_0-y_0)/m_0,0)$
$p_n=(x_0+m_0 y_0,0)$

and their distance

$norm(p_n-p_t)=(b^2costheta_0+a^2sintheta_0tantheta_0)/a$

but $norm(p_n-p_t)=a$ so

$b^2costheta_0+a^2sintheta_0tantheta_0=a^2$

or

$b^2cos^2theta_0+a^2sin^2theta_0=a^2costheta_0$

or

$(b^2/a^2-1)cos^2theta_0-cos theta_0+1=0$

or

$e^2 cos^2theta_0-cos theta_0+1=0$