InterviewSolution
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InterviewSolution
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Show that the midpoints of focal chords of a hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1`lie on another similar hyperbola. |
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Answer» Let the chord AB be bisected at point `P(h,k).` So, equation of chord AB is `(hx)/(a^(2))-(ky)/(b^(2))=(h^(2))/(a^(2))-(k^(2))/(b^(2))" (Using T = S"_(1)")"` Let it pass through the focus (ae, 0). (1) `therefore" "(he)/(a)=(h^(2))/(a^(2))-(k^(2))/(b^(2))` Therefore, locus of point P is (2) `(x^(2))/(a^(2))-(y^(2))/(b^(2))=(ex)/(a)` `rArr" "(1)/(a^(2))[x^(2)-aex]-(y^(2))/(b^(2))=0` `rArr" "((x-(ae)/(2))^(2))/(a^(2))-(y^(2))/(b^(2))=(e^(2))/(4)` This is also hyperbola with eccentricity e. |
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