The length of the chord of the ellipse frac{x | vedclass

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(D) Given the ellipse equation $\frac{x^2}{4} + y^2 = 1$ and the line $y = x + 1$. Substitute $y = x + 1$ into the ellipse equation: $\frac{x^2}{4} +

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$\frac{8}{5} \sqrt{2}$

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(D) Given the ellipse equation $\frac{x^2}{4} + y^2 = 1$ and the line $y = x + 1$. Substitute $y = x + 1$ into the ellipse equation: $\frac{x^2}{4} + (x + 1)^2 = 1$ $\frac{x^2}{4} + x^2 + 2x + 1 = 1$ $\frac{5x^2}{4} + 2x = 0$ $x(\frac{5x}{4} + 2) = 0$ So,$x_1 = 0$ and $x_2 = -\frac{8}{5}$. Corresponding $y$ values are $y_1 = 0 + 1 = 1$ and $y_2 = -\frac{8}{5} + 1 = -\frac{3}{5}$. The points of intersection are $P(0, 1)$ and $Q(-\frac{8}{5}, -\frac{3}{5})$. The length of the chord $PQ$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Length $= \sqrt{(-\frac{8}{5} - 0)^2 + (-\frac{3}{5} - 1)^2}$ $= \sqrt{(-\frac{8}{5})^2 + (-\frac{8}{5})^2} = \sqrt{\frac{64}{25} + \frac{64}{25}} = \sqrt{\frac{128}{25}} = \frac{8\sqrt{2}}{5}$.

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 $(3, 0)$	$(3)$ $x = 3$
	$(4)$ $y + 4 = 0$
	$(5)$ $x + 3 = 0$
	$(6)$ $3x + 25 = 0$

The length of the chord of the ellipse $\frac{x^2}{4} + y^2 = 1$ formed on the line $y = x + 1$ is

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