Question

Chords of an ellipse are drawn through the positive end of the minor axis. Their midpoint lies on

• A. a circle
• B. a parabola
• C. an ellipse
• D. a hyperbola

Answer 1

The Correct Option is C

Let the standard equation of the ellipse be: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Let the chord be drawn through the point \( (0, b) \), which is the positive end of the minor axis. Let this chord intersect the ellipse at points \( A \) and \( B \), and let \( M \) be the midpoint of chord \( AB \). Let the coordinates of the midpoint \( M \) be \( (h, k) \). Since the chord passes through \( (0, b) \) and midpoint is \( (h, k) \), the other end of the chord is at: \[ 2h, 2k - b \] Now, since both endpoints of the chord lie on the ellipse, their coordinates must satisfy the ellipse equation. Substitute the second endpoint into the ellipse: \[ \frac{(2h)^2}{a^2} + \frac{(2k - b)^2}{b^2} = 1 \] Simplifying: \[ \frac{4h^2}{a^2} + \frac{(2k - b)^2}{b^2} = 1 \] This is the locus of the midpoint \( (h, k) \). Multiply through by appropriate factors: \[ \frac{4h^2}{a^2} + \frac{(2k - b)^2}{b^2} = 1 \] This equation represents the locus of the midpoints of all such chords, and hence, the locus is: \[ \boxed{\frac{4x^2}{a^2} + \frac{(2y - b)^2}{b^2} = 1} \] where \( (x, y) \) are the coordinates of the midpoint. 

So, the midpoint of all such chords lies on a curve which is an ellipse, shifted vertically.