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Show that the locus represented by `x=(1)/(2)a(t+(1)/(t)),y=(1)/(2)a(t-(1)/(t))` is a rectangular hyperbola. |
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Answer» Let point P be `(a sec theta, b tan theta)`. Equation of tangent at point P is `(x)/(a)sec theta-(y)/(b)tan theta=1` Equation of asymptotes are `y=pm(b)/(a)x.` Solving asymptotes with tangent, we get `Q-=((a)/(sectheta-tantheta),(b)/(sectheta-tantheta))` `"and "R-=((a)/(sectheta+tantheta),(-b)/(sectheta+tantheta))` `therefore" Area of triangle CQR"=(1)/(2)||(0,0),((a)/(sectheta-tantheta),(b)/(sectheta-tantheta)),((a)/(sectheta+tantheta),(-b)/(sectheta+tantheta)),(0,0)||` `=(1)/(2)|-(a)/(sectheta-tantheta).(b)/(sectheta+tantheta)-(a)/(sectheta-tantheta).(b)/(sectheta+tantheta)|` = ab, which is constant. Also, midpoint of QR is `(((1)/(sectheta-tantheta)+(1)/(sectheta+tantheta))/(2),((b)/(sectheta-tantheta)-(b)/(sectheta+tantheta))/(2))` `-=(a sec theta, b tan theta)`, which is point P. |
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