If P_1 and P_2 are two points on the ellipse | vedclass

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(A) The slope of the chord joining $(0, 1)$ and $(2, 0)$ is $m = \frac{0-1}{2-0} = -\frac{1}{2}$.Since the tangents are parallel to this chord,their s

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$\sqrt{10}$

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(A) The slope of the chord joining $(0, 1)$ and $(2, 0)$ is $m = \frac{0-1}{2-0} = -\frac{1}{2}$.Since the tangents are parallel to this chord,their slope is $m = -\frac{1}{2}$.The equation of a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.Here $a^2 = 4$,$b^2 = 1$,and $m = -\frac{1}{2}$.So,$y = -\frac{1}{2}x \pm \sqrt{4(-\frac{1}{2})^2 + 1} = -\frac{1}{2}x \pm \sqrt{1+1} = -\frac{1}{2}x \pm \sqrt{2}$.The points of contact for the tangent $y = mx + c$ are $(\frac{-a^2m}{c}, \frac{b^2}{c})$.For $y = -\frac{1}{2}x + \sqrt{2}$,$P_1 = (\frac{-4(-1/2)}{\sqrt{2}}, \frac{1}{\sqrt{2}}) = (\sqrt{2}, \frac{1}{\sqrt{2}})$.For $y = -\frac{1}{2}x - \sqrt{2}$,$P_2 = (\frac{-4(-1/2)}{-\sqrt{2}}, \frac{1}{-\sqrt{2}}) = (-\sqrt{2}, -\frac{1}{\sqrt{2}})$.The distance between $P_1$ and $P_2$ is $\sqrt{(\sqrt{2} - (-\sqrt{2}))^2 + (\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}))^2} = \sqrt{(2\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{8 + 2} = \sqrt{10}$.

If $P_1$ and $P_2$ are two points on the ellipse $\frac{x^2}{4} + y^2 = 1$ such that the tangents at these points are parallel to the chord joining the points $(0, 1)$ and $(2, 0)$,then the distance between $P_1$ and $P_2$ is:

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