If S and S^{prime} are the foci of an ellipse | vedclass

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(A) For the ellipse $\frac{x^2}{169} + \frac{y^2}{144} = 1$,we have $a^2 = 169$ and $b^2 = 144$,so $a = 13$ and $b = 12$. The eccentricity $e$ is give

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$\left(0, \frac{10}{3}\right)$

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(A) For the ellipse $\frac{x^2}{169} + \frac{y^2}{144} = 1$,we have $a^2 = 169$ and $b^2 = 144$,so $a = 13$ and $b = 12$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$. The foci are $S = (ae, 0) = (13 	imes \frac{5}{13}, 0) = (5, 0)$ and $S^{\prime} = (-ae, 0) = (-5, 0)$. The point $B$ on the positive $Y$-axis is $(0, b) = (0, 12)$. The triangle $SBS^{\prime}$ has vertices $S(5, 0)$,$S^{\prime}(-5, 0)$,and $B(0, 12)$. The lengths of the sides are $SS^{\prime} = 10$,$BS = \sqrt{(5-0)^2 + (0-12)^2} = \sqrt{25 + 144} = 13$,and $BS^{\prime} = \sqrt{(-5-0)^2 + (0-12)^2} = \sqrt{25 + 144} = 13$. The incenter $(x, y)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and opposite side lengths $a, b, c$ is $(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c})$. Here,$x_1=5, y_1=0$ (opposite side $a=13$),$x_2=-5, y_2=0$ (opposite side $b=13$),$x_3=0, y_3=12$ (opposite side $c=10$). $x = \frac{13(5) + 13(-5) + 10(0)}{13+13+10} = 0$. $y = \frac{13(0) + 13(0) + 10(12)}{36} = \frac{120}{36} = \frac{10}{3}$. Thus,the incenter is $(0, \frac{10}{3})$.

List-$I$	List-$II$
$(i)$ The equation of the major axis	$(p)$ $3x = 34$
$(ii)$ The equation of a directrix	$(q)$ $y = 2$
$(iii)$ The equation of a latus rectum	$(r)$ $x + y = 9$
	$(s)$ $x = 6$
	$(t)$ $x = 3$
	$(u)$ $3y = 34$

If $S$ and $S^{\prime}$ are the foci of an ellipse $\frac{x^2}{169}+\frac{y^2}{144}=1$ and the point $B$ lying on the positive $Y$-axis is one end of its minor axis,then the incenter of the triangle $SBS^{\prime}$ is

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