Question

what is the maximum no. of common chords between a hyperbola and an ellipse?

Answer 1

Given :

Two conic sections hyperbola and ellipse .

To find :

Maximum no. of common chords.

Solution:

We know that,

Equation of ellipse :

$\\ \bf \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1$

Equation of hyperbola :

$\\ \bf \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1$

Maximum number of points of intersection of ellipse and hyperbola can be 4 as we can see in given figure :

For making a common chord, we will choose any 2 points from given 4 points,

For choosing r elements from total of n elements, combination formula is used,

So , the formula of combination is

$\\ \bf = \binom{n}{r} = ^nC_r = \frac{n!}{(n - r)!r!}$

In this case, there is total 4 common points of two given conic sections, so n = 4, in which 2 points is to be choosen to make a chord, so r = 2.

So,

Total number of common chords :

$\\ \bf = \binom{4}{2} = ^4C_2 = \frac{4!}{(4 - 2)!2!} \\ \bf = \frac{4!}{2!2!} = \frac{(4 \times 3 \times 2 \times 1)}{(2 \times 1)(2 \times 1)} = 6$

So, total 6 common chords can be formed by intersection of hyperbola and ellipse.