Question

For an ellipse, the locus of mid points of chords which are drawn through an end of minor axis will be

• A. a parabola
• B. an ellipse
• C. a hyperbola
• D. a pair of lines

Answer 1

Answer: (B)

Let the equation of the given ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .
Let $\left(acos\theta ,bsin\theta \right)$ be the co-ordinates of the other extremities of the chord of the ellipse.
So, the positive end of the minor axis is $\left(0,b\right)$ .
Let $\left(h,k\right)$ be the mid-point of the chord.
So, $h=\frac{acos\theta +0}{2}\Rightarrow cos\theta =\frac{2h}{a}..........\left(i\right)$
and $k=\frac{bsin\theta +b}{2}\Rightarrow sin\theta =\frac{2k-b}{b}..........\left(ii\right)$
Now, squaring and adding both the equations, we get
${\left(sin\right)}^2\theta +{\left(cos\right)}^2\theta ={\left(\frac{2k-b}{b}\right)}^2+{\left(\frac{2h}{a}\right)}^2$
$\Rightarrow \frac{4k^2+b^2-4kb}{b^2}+\frac{4h^2}{a^2}=1$
$\Rightarrow \frac{4k^2-4kb}{b^2}+1+\frac{4h^2}{a^2}=1$
$\Rightarrow \frac{4k^2-4kb}{b^2}+\frac{4h^2}{a^2}=0$
$\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}-\frac{k}{b}=0$
For locus, replacing $h\to x\&k\to y$ , we get
$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{y}{b}=0$ which is an ellipse
So, the required locus is an ellipse.