Let the ellipse E : x^2 + 9y^2 = 9 intersect | vedclass

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(C) The ellipse is $E : \frac{x^2}{9} + \frac{y^2}{1} = 1$. The points of intersection with the positive axes are $A(3, 0)$ and $B(0, 1)$.The major ax

Vedclass · Accepted Answer

$17$

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(C) The ellipse is $E : \frac{x^2}{9} + \frac{y^2}{1} = 1$. The points of intersection with the positive axes are $A(3, 0)$ and $B(0, 1)$.The major axis of $E$ lies along the $x$-axis with length $2a = 6$. Thus,the circle $C$ with the major axis as diameter has center $(0, 0)$ and radius $r = 3$. The equation of circle $C$ is $x^2 + y^2 = 9$.The line passing through $A(3, 0)$ and $B(0, 1)$ has the equation $\frac{x}{3} + \frac{y}{1} = 1$,or $x + 3y = 3$. Substituting $x = 3 - 3y$ into the circle equation $x^2 + y^2 = 9$:$(3 - 3y)^2 + y^2 = 9$$9 - 18y + 9y^2 + y^2 = 9$$10y^2 - 18y = 0$$2y(5y - 9) = 0$So,$y = 0$ (which corresponds to point $A$) or $y = \frac{9}{5}$.If $y = \frac{9}{5}$,then $x = 3 - 3(\frac{9}{5}) = 3 - \frac{27}{5} = -\frac{12}{5}$. Thus,$P = (-\frac{12}{5}, \frac{9}{5})$.The area of triangle $OAP$ with vertices $O(0, 0)$,$A(3, 0)$,and $P(-\frac{12}{5}, \frac{9}{5})$ is given by $\frac{1}{2} |x_O(y_A - y_P) + x_A(y_P - y_O) + x_P(y_O - y_A)| = \frac{1}{2} |0 + 3(\frac{9}{5} - 0) + (-\frac{12}{5})(0 - 0)| = \frac{1}{2} |\frac{27}{5}| = \frac{27}{10}$.Here $m = 27$ and $n = 10$,which are coprime. Therefore,$m - n = 27 - 10 = 17$.

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