Let x = 9 be a directrix of an ellipse E, who | vedclass

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(A) Given the directrix $x = a/e = 9$ and eccentricity $e = 1/3$, we find $a = 9 	imes (1/3) = 3$.
The focus $P$ is at $(ae, 0) = (3 	imes 1/3, 0)

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$9y^2 = 8x(1 - x)$

Vedclass · Answer

(A) Given the directrix $x = a/e = 9$ and eccentricity $e = 1/3$, we find $a = 9 	imes (1/3) = 3$.
The focus $P$ is at $(ae, 0) = (3 	imes 1/3, 0) = (1, 0)$, so $\alpha = 1$.
The equation of the ellipse is $x^2/a^2 + y^2/b^2 = 1$, where $b^2 = a^2(1 - e^2) = 9(1 - 1/9) = 8$.
Thus, the ellipse is $x^2/9 + y^2/8 = 1$.
The locus of the midpoint $(h, k)$ of a chord passing through a point $(x_0, y_0)$ for an ellipse $x^2/a^2 + y^2/b^2 = 1$ is given by $T = S_1$, which is $xh/a^2 + yk/b^2 = h^2/a^2 + k^2/b^2$.
Since the chord passes through $(1, 0)$, the equation of the chord with midpoint $(h, k)$ is $xh/9 + yk/8 = h^2/9 + k^2/8$.
Substituting $(x, y) = (1, 0)$ into this equation, we get $h/9 = h^2/9 + k^2/8$.
Multiplying by $72$, we get $8h = 8h^2 + 9k^2$, which simplifies to $9k^2 = 8h(1 - h)$.
Replacing $(h, k)$ with $(x, y)$, the locus is $9y^2 = 8x(1 - x)$.

Let $x = 9$ be a directrix of an ellipse $E$, whose centre is at the origin and eccentricity is $1/3$. Let $P(\alpha, 0), \alpha > 0$, be a focus of $E$ and $AB$ be a chord passing through $P$. Then the locus of the mid point of $AB$ is :

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