Question

Variable pairs of chords at right angles are drawn through any point P on the ellipse $\frac{x^2}{4} + y^2 = 1$, to meet the ellipse at two points say A and B. If the line joining A and B passes through a fixed point Q(a, b) such that $a^2 + b^2$ has the value equal to $\frac{m}{n}$, where m, n are relatively prime positive integers, find (m + n).

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (B)

6

Answer 2

The locus of the point P from which perpendicular chords are drawn to an ellipse is its director circle. For the ellipse $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$, the director circle is $x^2 + y^2 = \alpha^2 + \beta^2$. For the given ellipse $\frac{x^2}{4} + y^2 = 1$, we have $\alpha^2 = 4$ and $\beta^2 = 1$. Thus, the director circle is $x^2 + y^2 = 4 + 1 = 5$. A property of ellipses states that if two perpendicular chords are drawn through a point P, the locus of the foot of the perpendicular from the center of the ellipse to the chord joining the extremities (A and B) of these chords is the director circle. Since the line AB passes through the fixed point Q(a, b), the point Q must lie on the director circle. Therefore, $a^2 + b^2 = 5$. We are given $a^2 + b^2 = \frac{m}{n}$. Comparing, we get $\frac{m}{n} = 5 = \frac{5}{1}$. So, $m=5$ and $n=1$. These are relatively prime positive integers. The required value is $m+n = 5+1 = 6$.