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Find the equation of the ellipse whose focus is `S(-1, 1),` the corresponding directrix is `x -y+3=0,` and eccentricity is 1/2. Also find its center, the second focus, the equation of the second directrix, and the length of latus rectum. |
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Answer» y-1=-1(x+1) y=-x CM=a/e=2a CS=ae=a/2 SM=CM-CS=2a-a/2=3a/2 SM=`|(-1-1+3)/sqrt2|=3a/2` after solving this a=`sqrt2/3` `a^2=2/9` `e^2=1-b^2/a^2=1/2` `b^2/a^2=3/4` `b^2=1/6` CM=2a `|(2alpha+3)/sqrt2|=2xsqrt2/3` solving this we get `alpha=-5/6` now, we can find latus rectum latus rectum=`(2b^2)/a=1/sqrt2` `sqrt((h+1)^2+(h+1)^2)=sqrt2/3` h=-2/3 `s=(-2/3,2/3)` `D_1: x-y+3=0` `D_2: x-y+lambda=0` distance between `D_1 and D_2` `|(alpha-3)/sqrt2|=4sqrt2/3` `alpha=1/3 and 17/3` 17/3 is not possible `d_2:x-y+1/3=0` |
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