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

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Question

Any tangent on an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ given by

$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $

Here&nb

Vedclass · Accepted Answer

$\sqrt {10} $

Vedclass · Answer

Any tangent on an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ given by

$y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $

Here $a = 2,b = 1$

$m = \frac{{1 - 0}}{{0 - 2}} = \frac{1}{2}$

$c = \sqrt {4{{\left( { - \frac{1}{2}} ight)}^2} + {1^2}}  = \sqrt 2 $

So,$y =  - \frac{1}{2}x \pm \sqrt 2 $

For ellipse $:\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$

We put $y =  - \frac{1}{2}x \pm \sqrt 2 $

$	herefore \frac{{{x^2}}}{4} + {\left( { - \frac{x}{2} + \sqrt 2 } ight)^2} = 1$

$\frac{{{x^2}}}{4} + \left( {\frac{{{x^2}}}{4} - 2\left( {\frac{x}{2}} ight)\sqrt 2  + 2} ight) = 1$

$ \Rightarrow {x^2} + 2\sqrt 2 x + 2 = 0$

or ${x^2} - 2\sqrt 2 x + 2 = 0$

$ \Rightarrow x = \sqrt 2 $ or $ - \sqrt 2 $

If $x = \sqrt 2 ,y = \frac{1}{{\sqrt 2 }}$ and $x =  - \sqrt 2 ,y =  - \frac{1}{{\sqrt 2 }}\,$

$	herefore $ Points are $\left( {\sqrt 2 ,\frac{1}{{\sqrt 2 }}} ight),\left( { - \sqrt 2 , - \frac{1}{{\sqrt 2 }}} ight)\,$

$	herefore {P_1}{P_2} = \sqrt {{{\left\{ {\frac{1}{{\sqrt 2 }} - \left( {\frac{1}{{\sqrt 2 }}} ight)} ight\}}^2} + {{\left\{ {\sqrt 2  - \left( { - \sqrt 2 } ight)} ight\}}^2}} \,$

$ = \sqrt {{{\left( {\frac{2}{{\sqrt 2 }}} ight)}^2} + {{\left( {2\sqrt 2 } ight)}^2}}  = \sqrt {2 + 8}  = \sqrt {10} $

If $P_1$ and $P_2$ are two points on the ellipse $\frac{{{x^2}}}{4} + {y^2} = 1$ at which the tangents 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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