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The gradient of one of the lines given by `ax^(2)+2hxy+by^(2)=0` is twice that of the other, thenA. `h^(2)=ab`B. `h=a+b`C. `8h^(2)=9ab`D. `9h^(2)=8ab` |
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Answer» Correct Answer - C Let m and be the gradients of the lines given by `ax^(2)+2hxy+by^(2)=0`. Then, `m+2m=(-2h)/(b)and mxx2m=(a)/(b)` `rArr" "m=-(2h)/(3b)and m^(2)=(a)/(2b)` `rArr" "(4h^(2))/(9b^(2))=(a)/(2b)=rArr 8h^(2)=9ab` |
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