Question I:
Given the base of a triangle and the ratio of the tangents of half the base angles, prove that the vertex moves on a hyperbola, whose foci are the extremities of the base.
Solution I:
Let the base of the triangle ABC be , and
Then, we get
$\Longrightarrow latex a+(b-c)=K(a-(b-c))$
Since a and K are given constants, is a constant, say
. Therefore, we get
, that is,
. We find that the vertex A moves in such a manner that the difference of its distances from two fixed points B and C, which are the extremities of the base is constant. Hence, it describes a hyperbola whose foci are the extremities of the base. For instance, you will get further insight if you try the following: (in previous blog, basics 4, of coordinate geometry) : Show that the difference of focal distances of any point on a hyperbola is constant and equal to the length of the transverse axis.
Question 2:
Show that the angle between the tangents to the ellipse and the circle
at their points of intersection is
.
Solution 2:
For the points of intersection, we have
Consider the point , intersection of the given ellipse and the circle. Equation of the tangent at P to the circle is
Slope of this tangent is . Equation of the tangent at P to the ellipse is
and slope of this tangent is
. If
be the angle between these tangents, then
.
Hence, . Note: The angle will be the same at each point of intersection.
Question 3:
A straight line is drawn parallel to the conjugate axis of the hyperbola to meet it and the conjugate hyperbola in the points P and Q respectively. Show that the normals at P and Q to the curves meet on the x-axis.
Solution 3:
Conjugate Hyperbola of the given hyperbola is the hyperbola whose transverse and conjugate axis are respectively the conjugate and transverse axis of the given hyperbola and thus the equation is .
Now, is a point of the given hyperbola and
is a point on the conjugate hyperbola. Since P and Q lie on the line parallel to the conjugate axis
, of the given hyperbola,
,
.
Now, equation of the normal at P to the given hyperbola is …call this equation I.
and equation of the normal at Q to the conjugate hyperbola is ….call this equation II.
Eliminating x from I and II, using we get:
, which in turn, implies that
. Hence, the normals meet on the x-axis.
More later, including homeworks,
Nalin Pithwa.