InterviewSolution
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InterviewSolution
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If a variable straight line `x cos alpha+y sin alpha=p` which is a chord of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 (b gt a)` subtends a right angle at the centre of the hyperbola, then it always touches a fixed circle whose radius, isA. `(ab)/(sqrt(b-2a))`B. `(a)/(a-b)`C. `(ab)/(sqrt(b^(2)-a^(2)))`D. `(ab)/(sqrt(b+a))` |
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Answer» The combined equation of the straight lines joining, the centre of the hyperbola of the points of intersection of the line `x cosalpha+ysinalpha=p` and the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is `(x^(2))/(a^(2))-(y^(2))/(b^(2))=((xcosalpha+ysinalpha)/(p))^(2)` This is equation will represent a pair of perpendicular straight lines, if `(1)/(a^(2))-(cos^(2)alpha)/(p^(2))-(1)/(b^(2))-(sin^(2)alpha)/(p^(2))=0impliesp=(ab)/(sqrt(b^(2)-a^(2)))` Substituting the value of `p` in `x cosalpha+ysinalpha=p`, we get `x cosalpha+ysinalpha=(ab)/(sqrt(b^(2)-a^(2)))` Clearly, it touches the circle `x^(2)+y^(2)=(-(ab)/(sqrt(b^(2))-a^(2))^(2))` |
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