Question

Let \( A_1 \) be the area of the given ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Let \( A_2 \) be the area of the region bounded by the curve which is the locus of mid point of the line segment joining the focus of the ellipse and a point P on the given ellipse, then \( A_1 : A_2 = \)

• A. 3:2
• B. a:b
• C. 4:1
• D. 2a:3b

Hint:

Area of ellipse \( \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \) is \( \pi AB \). To find the locus of the midpoint of a segment joining a fixed point \( (x_f, y_f) \) and a variable point \( (x_0, y_0) \) on a curve, let the midpoint be \( (h,k) \). Then \( h = (x_f+x_0)/2 \) and \( k = (y_f+y_0)/2 \). Express \( x_0, y_0 \) in terms of \( h,k \) and substitute into the curve's equation. The locus of the midpoint of chords from a point P on an ellipse to a focus S is another ellipse scaled by a factor of 1/2.

Answer 1

The Correct Option is C

The area of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( A_1 = \pi ab \).
Let one focus of the ellipse be \( S = (ae, 0) \) (assuming \( a > b \)).
Let \( e \) be the eccentricity.
Let \( P \) be a point \( (x_0, y_0) \) on the ellipse, so \( \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1 \).
Let \( M(h, k) \) be the midpoint of the line segment \( SP \).
\[ h = \frac{ae + x_0}{2} \implies x_0 = 2h - ae \] \[ k = \frac{0 + y_0}{2} \implies y_0 = 2k \] Substitute \( x_0 \) and \( y_0 \) into the equation of the ellipse: \[ \frac{(2h - ae)^2}{a^2} + \frac{(2k)^2}{b^2} = 1 \] \[ \frac{4\left(h - \frac{ae}{2}\right)^2}{a^2} + \frac{4k^2}{b^2} = 1 \] Divide by 4: \[ \frac{\left(h - \frac{ae}{2}\right)^2}{a^2/4} + \frac{k^2}{b^2/4} = 1 \] The locus of \( M(h, k) \) is an ellipse: \[ \frac{(x - \frac{ae}{2})^2}{(a/2)^2} + \frac{y^2}{(b/2)^2} = 1 \] This is an ellipse with semi-major axis \( a' = \frac{a}{2} \) and semi-minor axis \( b' = \frac{b}{2} \).
The centre of this new ellipse is \( \left(\frac{ae}{2}, 0\right) \).
The area of this new ellipse (locus of \( M \)) is: \[ A_2 = \pi a' b' = \pi \left(\frac{a}{2}\right) \left(\frac{b}{2}\right) = \frac{\pi ab}{4} \] We need the ratio \( A_1 : A_2 \).
\[ A_1 : A_2 = \pi ab : \frac{\pi ab}{4} \] \[ A_1 : A_2 = 1 : \frac{1}{4} \] \[ A_1 : A_2 = 4 : 1 \] This matches option (3).
The choice of focus \( (-ae, 0) \) or if \( b > a \) focus \( (0, be) \) would lead to a similar ellipse, just shifted, with the same semi-axes \( a/2, b/2 \).