251. Problems of Analytical Geometry.
by Virgil Nicula, Mar 11, 2011, 7:03 AM
PP1. Let
be a circle and let
be a line which is tangent to
in the fixed point
. Consider two mobile
points
so that
, where
is constant and
. Denote the intersection 
of the tangents (different from
) from
,
to
. Prove that the geometrical locus of
is a line
which is parallelly with
.
Proof. Choose one system of coordinates
so that the equation of the circle
is
and the equation of the line
is
, i.e.
.
Consider two mobile points
,
for which
(constant) and
. Show easily that the slopes
,
of the tangents
,
from
,
respectively (different from
) to
are
and the equations of the tangents
, 
are
.
In conclusion,
, i.e. the geometrical locus of the mobile point
is a line
for which
. See second and third problems from here.
See here and here another problems of analytical geometry.
PP2. Ascertain
for which the equation
has at least a real root.
Proof. For one fixed
the relation
is the equation
of a line
in the analytic system
. The distance
from
origin
to
must be equally to the distance
from the origin to the line
. Thus,
.
The minimum value of the function
,
is
. In conclusion,
.
PP3. Show that if the equation
has at least a real root, then
.
Proof. Consider the equation
of the line
in the analytical system
, where
. The distance
from the origin 
to
is equally to the distance
from the origin to the line
. Thus,
. In conclusion,
.
Remark. Clearly
. Thus,
.
PP4. Let
be an ellipse with foci
and
. Consider for a point
the tangent line
and the normal line 
through the point
to the given ellipse. Prove that normal line
is the angle bisector of the angle
.
Proof. Let
be the equation of the ellipse
with the foci
,
, where
. Let
, where
.
Observe that
the slope of the tangent line to the ellipse in the point
is
and the slope of the normal line to the ellipse in the point
is
. Therefore, the equation of the tangent line
is 
and the equation of the normal line
is
. If denote
and
, then prove easily that
and
.
Observe that
, what is truly.
In conclusion, the division
is harmonically and in this case
.
PP5. Let
be a square and let
be its center. If
is a point lying on the segment
, prove that there exists the points
,
lying on the segments
and
respectively such that
is the orthocenter of
if and only if
.
Proof. Choose
,
,
,
and
,
,
, where
,
,
.
Therefore,
is the orthocenter of
.
Since
and
, from
,
obtain that
.
Remark.
.
PP6. Find the maximum area of the triangle formed by the intersection points between an ellipse
with the line passing through one of the foci and the other focus.
Proof. Given the ellipse
with foci
and
. Let
be the equation of a line
through
. Can suppose w.l.o.g. by
symmetry w.r.t. the
-axis that
and the line
cuts the ellipse in
and
, where
. Solving the system
obtain that
. Since
obtain that
. Since

obtain that
. Since
(constant) ![$[AMN]=c\cdot \left(y_m-y_n\right)$](//latex.artofproblemsolving.com/2/3/9/23933c6e34ca4403be0d67325982a231462dfcf2.png)
is
is
is
and in this case
. Prove easily that
. In conclusion,
.
Remark. I"ll use the notation
, i.e.
and
have Same Sign.
PP7. If
, then find range of
.
Proof 1. Using the geometric interpretation, it means finding the tangency points between the circle centred at
of radius
and the family of equilateral
hyperbolae
. Because of the symmetry of the configuration, these occur at points on the first bisectrix
. Thus at
,
therefore
.
Proof 2. Using the substitutions
, where
obtain that
, i.e.
, where
and
. Since
prove easily that the range of the function ![$f:\left[-\sqrt 2,\sqrt 2\right]\rightarrow \mathbb R$](//latex.artofproblemsolving.com/7/a/a/7aac8260de8a3e3fe3e5504e3f963c8531ba5d9d.png)
is
, i.e.
, i.e.
.




points





of the tangents (different from







Proof. Choose one system of coordinates






Consider two mobile points















are



In conclusion,




See here and here another problems of analytical geometry.
PP2. Ascertain


Proof. For one fixed






origin





The minimum value of the function




PP3. Show that if the equation


Proof. Consider the equation






to





Remark. Clearly




PP4. Let






through the point



Proof. Let







Observe that












and the equation of the normal line






Observe that


In conclusion, the division



PP5. Let



![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)


![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)



Proof. Choose










Therefore,




Since







Remark.


![$p\in\left[\frac 13,1\right]$](http://latex.artofproblemsolving.com/5/6/c/56c3c44d2e6ab0925b5b3f997308fba2248d9fc3.png)
PP6. Find the maximum area of the triangle formed by the intersection points between an ellipse

Proof. Given the ellipse






symmetry w.r.t. the





















![$[AMN]=c\cdot \left(y_m-y_n\right)$](http://latex.artofproblemsolving.com/2/3/9/23933c6e34ca4403be0d67325982a231462dfcf2.png)
is





![$[AMN]=2acb^2\cdot f(\lambda )$](http://latex.artofproblemsolving.com/d/a/d/dad223744d411661983b3c2a05927bc09bbfd88d.png)

![$\left|\begin{array}{c}
b<a\le b\sqrt 2\ \implies\ f(\lambda )\le f(\infty )=\frac {1}{a^2}\ \implies\ [AMN]\le\frac {2cb^2}{a}\\\\
b\sqrt 2<a\ \implies\ f(\lambda )\le f\left(\frac {b}{\sqrt{a^2-2b^2}}\right)=\frac {1}{2bc}\ \implies\ [AMN]\le ab\end{array}\right|$](http://latex.artofproblemsolving.com/a/4/c/a4c21f8d7cc3b1eefb396abf7ee685524942a1d8.png)
![$\implies [AMN]\le ab$](http://latex.artofproblemsolving.com/b/8/6/b86944b742ebf13514db74089cf428fe8624eca8.png)




PP7. If


Proof 1. Using the geometric interpretation, it means finding the tangency points between the circle centred at


hyperbolae



therefore

Proof 2. Using the substitutions

![$\phi\in\left[0,2\pi\right]$](http://latex.artofproblemsolving.com/f/c/8/fc8322641a75e1af4e23ce0431c2ad450afd8102.png)





![$f:\left[-\sqrt 2,\sqrt 2\right]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/7/a/a/7aac8260de8a3e3fe3e5504e3f963c8531ba5d9d.png)
is
![$\left[f\left(-\sqrt 2\right), f\left(\sqrt 2\right)\right]$](http://latex.artofproblemsolving.com/e/9/c/e9cf916205053f923a52a8ba4495bd6191f96356.png)
![$\left[11-6\sqrt 2, 11+6\sqrt 2\right]$](http://latex.artofproblemsolving.com/7/d/f/7df22890958b60fddb11819e0147f7e650aed424.png)

This post has been edited 101 times. Last edited by Virgil Nicula, Nov 22, 2015, 11:07 AM