For the ellipse frac{x^2}{25}+frac{y^2}{16}=1 | vedclass

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(P-6, Q-1, R-3) The given equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here,$a^2=25 \Rightarrow a=5$ and $b^2=16 \Rightarrow b=4$. Th

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(P-6, Q-1, R-3) The given equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here,$a^2=25 \Rightarrow a=5$ and $b^2=16 \Rightarrow b=4$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.$(P)$ The directrix corresponding to the focus $(-ae, 0) = (-3, 0)$ is $x = -\frac{a}{e} = -\frac{5}{3/5} = -\frac{25}{3}$,which gives $3x + 25 = 0$. This matches $(6)$.$(Q)$ The tangent at the vertex $(0, 4)$ is $y = 4$. This matches $(1)$.$(R)$ The latus rectum through the focus $(ae, 0) = (3, 0)$ is $x = ae = 5 	imes \frac{3}{5} = 3$. This matches $(3)$.

List-$I$	List-$II$
$(P)$ Directrix corresponding to the focus $(-3, 0)$	$(1)$ $y = 4$
$(Q)$ Tangent at the vertex $(0, 4)$	$(2)$ $3x = 25$
$(R)$ Latus rectum through $(3, 0)$	$(3)$ $x = 3$
	$(4)$ $y + 4 = 0$
	$(5)$ $x + 3 = 0$
	$(6)$ $3x + 25 = 0$

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