Geo5.2.18

From Example Problems
Jump to: navigation, search

Show that the foot of the perpendicular drawn from the centre on any tangent to the ellipse lies on the curve (x^{2}+y^{2})^{2}=a^{2}x^{2}+b^{2}y^{2}\,

The equation of the tangent at any point P(\theta )=(a\cos \theta ,b\sin \theta )\, is

{\frac  {x\cos \theta }{a}}+{\frac  {y\cos \theta }{b}}=1\, Equation 1.

Let M(x_{1},y_{1})\, be the foot of the perpendicular from the centre (0,0) to the

tangent,then Product of the slopes of PM and CM is -1.

Hence

{\frac  {y_{1}}{x_{1}}}\cdot {\frac  {{\frac  {-\cos \theta }{a}}}{{\frac  {\sin \theta }{b}}}}=-1\,

{\frac  {x_{1}}{{\frac  {\cos \theta }{a}}}}={\frac  {y_{1}}{{\frac  {\sin \theta }{b}}}}=\lambda \, (say)

ax_{1}=\lambda \cos \theta ,by_{1}=\lambda \sin \theta \,

\lambda ^{2}=a^{2}x_{1}^{{2}}+b^{2}y_{1}^{{2}}\,. Equation 2.

\lambda ={\sqrt  {a^{2}x_{1}^{{2}}+b^{2}y_{1}^{{2}}}}\,

Therefore, {\frac  {\cos \theta }{a}}={\frac  {x_{1}}{\lambda }},{\frac  {\sin \theta }{b}}={\frac  {y_{1}}{\lambda }}\, Equation 3.

But M lies on the first line.

From 3, {\frac  {x_{1}^{{2}}}{\lambda }}+{\frac  {y_{1}^{{2}}}{\lambda }}=1\,

Therefore, from 2,[x_{1}^{{2}}+y_{1}^{{2}}]^{2}=\lambda ^{2}=a^{2}x_{1}^{{2}}+b^{2}y_{1}^{{2}}\,

Henc the P lies on (x^{2}+y^{2})^{2}=(a^{2}x^{2}+b^{2}y^{2})\,


Main Page:Geometry:The Ellipse