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If two distinct chords, drawn from the point (p, q) on the circle `x^2+y^2=p x+q y`(where `p q!=q)`are bisected by the x-axis, then`p^2=q^2`(b) `p^2=8q^2``p^28q^2`A. `p^(2)=q^(2)`B. `p^(2)=8q^(2)`C. `p^(2)lt 8q^(2)`D. `p^(2)gt 8q^(2)` |
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Answer» Correct Answer - D Chord is bisected by x- axis , so that its mid point is (h,O) Hence by `T=s_(1)` its equation is `bx +0-(P)/(2)(y+0)= h^(2)-ph` it passes through the point (p,q) `therefore 2 h^(2)- 3ph+(p^(2)+q^(2))=0" ".....(1) ` it is a quadratic equation in h Since two distinct chords are drawn the roots of (1) will be real and distinct `therefore Dgt0implies9p^(2)-8(p^(2)+q^(2))gt 0` or ` p^(2) gt 8 q^(2)` |
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