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The sum of the squares of the perpendicular drawn from the points (0,1) and `(0,-1)` to any tangent to a curve is 2. The equation of the curve, isA. `2y=C(x+2)`B. `y=C(x+1)`C. `y=C(x+2)`D. `y=C(x+2)` |
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Answer» Correct Answer - B The equation of any tangent to a curve `y=f(x)` is `y-y=(dy)/(dx)(X-x)or, X(dy)/(dx)-Y+(y-x(dy)/(dx))=0" ...(i)"` It is given that `|(-1+y-x(dy)/(dx))/(sqrt(((dy)/(dx))^(2)+1))|^(2)+|(1+y-x(dy)/(dx))/(sqrt(((dy)/(dx))^(2)+1))|^(2)=2` `rArr" "2{(y-x(dy)/(dx))^(2)+1}=2{1+((dy)/(dx))^(2)}` `rArr" "(y-x(dy)/(dx))^(2)=((dy)/(dx))^(2)` `rArr" "y-x(dy)/(dx)=pm(dy)/(dx)` `rArr" "(1)/(xpm1)dx=(1)/(dy)dy` `rArr" "log(xpm1)=logy+logC` `rArr" "log(xpm1)=logy+logC` `rArr" "Cy=x pm1 or,y=k(x pm1)` |
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