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The tangents and normal at a point on (x^(2))/(a^(2))-(y^(2))/(b^(2)) =1 cut the y-axis A and B. Then the circle on AB as diameter passes through |
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Answer» ONE of the vertex of the HYPERBOLA `(sec theta)/(a) x -(tan theta)/(b) y =1` `:. A (0,-b cot theta)` Equation of NORMAL at point `P(theta)` is `a cos theta x + b cot theta y = a^(2) +b^(2)` `:. B(0,(a^(2)+b^(2))/(b cot theta))` Equation of circle as AB as a diameter is `(x-0) (x-0) + (y+b cot theta) (y-(a^(2)+b^(2))/(b cot theta)) =0` or `x^(2)+ (y+b cot theta) (y-(a^(2)e^(2))/(b cot theta)) =0` Clearly this passes through foci (ae,0) |
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