JEE Main & Advanced Sample Paper JEE Main Sample Paper-10

  • question_answer
    The line \[\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}\]intersect the curve \[xy={{c}^{2}},z=0\] if c is equal to

    A)  \[\pm \sqrt{5}\]                             

    B)  \[\pm \,4\]

    C)  \[\pm 1/2\]                      

    D)  \[\pm 1/\sqrt{3}\]

    Correct Answer: A

    Solution :

     Idea To find the points, the intersection of the given line and curve, put z = 0 in equation of line. These points lie on the curve xy = e2. For the points where the line intersects the curve, we have z = 0 \[\therefore \]                  \[\frac{x-2}{3}=\frac{y+1}{2}=\frac{0-1}{-1}\] \[\Rightarrow \]               \[\frac{x-2}{3}=1\]and\[\frac{y+1}{2}=1\] \[\Rightarrow \]               \[x=5,y=1\] Put\[x=5,y=1\]in \[xy={{c}^{2}}\]             \[c=\pm \sqrt{5}\] TEST Edge The intersection of line and curve and a line lie on the plane related question are asked. Students are suggested to understand the concept of line and plane.


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