**Problem 1:**

Find the locus of the point of intersection of tangents to the ellipse , which are at right angles.

**Solution I:**

Any tangent to the ellipse is ….call this equation I.

Equation of the tangent perpendicular to this tangent is …call this equation II.

The locus of the intersection of tangent lines (i) and (ii) is obtained by eliminating m between these equations. So, squaring and adding them, we get

which is a circle with its centre at the centre of the ellipse and radius equal to the length of the line joining the ends of the major and minor axis. This circle is called the **director circle **of the ellipse.

**Problem II:**

A tangent to the ellipse meets the ellipse at P and Q. Prove that the tangents at P and Q of the ellipse at right angles.

**Solution II:**

Let the tangent at to the ellipse meet the ellipse at P and Q.

Let the tangents at P and Q to the second ellipse intersect at the point . Then, PQ is the chord of contact of the point with respect to ellipse two, and so its equation is

….call this “A”.

PQ is also the tangent at to the first ellipse and so the equation can be written as ….call this “B”.

Comparing “A” and “B”, we get

and

The locus of is , or , which is the director circle of the second ellipse.

Hence, the tangents at P and Q to the ellipse (ii) are at right angles (by the solution to the previous example).

**Problem 3:**

Let d be the perpendicular distance from the centre of the ellipse to the tangent drawn at a point P on the ellipse. If and are the two foci of the ellipse, then show that

**Solution 3:**

Equation of the tangent at the point on the given ellipse is . Thus,

We know

…call this equation I.

Also, , which in turn equals,

, that is,

that is,

Now, from I, we get ,

also,

Hence,

we will continue later,

Cheers,

Nalin Pithwa

### Like this:

Like Loading...

*Related*