The locus of the mid-points of the chords of | vedclass

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(C) Given the equation of the ellipse is $x^{2}+4y^{2}=4$.Dividing by $4$, we get $\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$.The positive end of the minor ax

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an ellipse with centre $\left(0, \frac{1}{2}\right),$ major axis $1$ and minor axis $\frac{1}{2}$

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(C) Given the equation of the ellipse is $x^{2}+4y^{2}=4$.Dividing by $4$, we get $\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$.The positive end of the minor axis is $B(0, 1)$.Let the mid-point of the chord $BP$ be $M(h, k)$, where $P(x, y)$ is a point on the ellipse.Then, $(h, k) = \left(\frac{0+x}{2}, \frac{1+y}{2}\right)$.This implies $h = \frac{x}{2} \Rightarrow x = 2h$ and $k = \frac{1+y}{2} \Rightarrow y = 2k-1$.Since $P(x, y)$ lies on the ellipse $x^{2}+4y^{2}=4$, substituting the values of $x$ and $y$ gives:$(2h)^{2} + 4(2k-1)^{2} = 4$$4h^{2} + 4(4k^{2} - 4k + 1) = 4$$4h^{2} + 16k^{2} - 16k + 4 = 4$$4h^{2} + 16k^{2} - 16k = 0$Dividing by $4$, we get $h^{2} + 4k^{2} - 4k = 0$.Replacing $(h, k)$ with $(x, y)$, the locus is $x^{2} + 4(y^{2} - y) = 0$.Completing the square: $x^{2} + 4(y^{2} - y + \frac{1}{4}) = 4(\frac{1}{4}) = 1$.$x^{2} + 4(y - \frac{1}{2})^{2} = 1$.Dividing by $1$, we get $\frac{x^{2}}{1} + \frac{(y - 1/2)^{2}}{1/4} = 1$.This is an ellipse with centre $(0, 1/2)$, semi-major axis $a=1$, and semi-minor axis $b=1/2$.

The locus of the mid-points of the chords of an ellipse $x^{2}+4y^{2}=4$ that are drawn from the positive end of the minor axis is

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