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Prove that the chords of constant of perpendicular tangents to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` touch another fixed ellipse `(x^(2))/(a^(4))+(y^(2))/(b^(4))=(1)/((a^(2)+b^(2)))` |
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Answer» We know, that the locus of the point of intersection of perpendicular tangents to the given ellipse is `x^(2)+y^(2)=a^(2)+b^(2)`. Any point on this circle can be taken as `P-=(sqrt(a^(2)+b^(2))cos theta,sqrt(a^(2)+b^(2))sin theta)` The equation of the chord of conatact of tangents from P is `(x)/(a^(2))sqrt(a^(2)+b^(2)) cos theta+(y)/(b^(2))sqrt(a^(2)+b^(2))sin theta=1` Let this line be a tangent be a tangne to the fixed ellipse `(x^(2))/(A^(2))+(y^(2))/(B^(2))=1` `:. (x)/(A) cos theta +(y)/(B) sin theta=1` Where `A=(a^(2))/(sqrt(a^(2)+b^(2))),B=(b^(2))/(sqrt(a^(2)+b^(2)))` Hence, the ellipe is `(x^(2))/(a^(4))+(y^(2))/(b^(4))=(1)/((a^(2)+b^(2)))` |
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