**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.

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