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?

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

5400 views around the world

You can reuse this answer
Creative Commons License

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#