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If the curves `ay+x^2=7` and `x^3=y` cut orthogonally at `(1, 1)` then `a=` (A) 1 (B) `-6` (C) 6 (D) `1/6`A. 1B. 0C. `-6`D. 6 |
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Answer» Correct Answer - D We have, `ay=x^(2)=7 "and" x^(3)=y` On differentiating w.r.t. x in both equations, we get `a.(dy)/(dx)+2x=0` and `3x^(2)=(dy)/(dx)` `rArr (dy)/(dx)_(1,1)=-2/a=m_(1)` and `(dy)/(dx)_(1,1)=3.1=3=m_(2)` Since, the curves cut orthogonally at (1,1). `therefore m_(1).m_(2)=-1` `rArr (-2/a).3=-1` `therefore a=6` |
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