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JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    The sum of the intercepts on the coordinate axes of the plane passing through the points\[(-2,-2,2)\] and containing the line joining the points and is? The differential equation representing the family of ellipse having foci either on the x-axis or on the y-axis, centre at the origin and passing through the points\[(1,-1,2)\] and \[(1,1,1)\]is? [JEE Main 16-4-2018]

    A)  12                              

    B)  -8

    C)  -4                               

    D)  4

    Correct Answer: C

    Solution :

     \[\left| \begin{matrix}    x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}}  \\    {{z}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}}  \\    {{x}_{3}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}}  \\ \end{matrix} \right|=0\]                 \[\left| \begin{matrix}    x+2 & y+2 & z-2  \\    1+2 & -1+2 & 2-2  \\    1+2 & 1+2 & 1-2  \\ \end{matrix} \right|=0\]                 \[\left| \begin{matrix}    x+2 & y+2 & z+2  \\    3 & 1 & 0  \\    3 & 3 & -1  \\ \end{matrix} \right|=0\]                 \[\Rightarrow \]\[(z-2)(9-3)-1(x+2-3y-6)=0\]                 \[\Rightarrow \]\[6z-12-x-2+3y+6=0\]                 \[\Rightarrow \]\[-x+3y+6z-8=0\]                 \[\Rightarrow \]\[\frac{x}{8}-\frac{3y}{8}-\frac{6z}{8}+\frac{8}{8}=0\]                 \[\Rightarrow \]\[\frac{x}{8}-\frac{y}{\frac{8}{3}}-\frac{z}{\frac{8}{6}}=1\]                 \[\Rightarrow \]\[\frac{x}{-8}+\frac{9}{\frac{8}{3}}+\frac{z}{\frac{8}{6}}=1\]                 Sum of intercepts                 \[=-8+\frac{8}{3}+\frac{8}{6}\]                 \[=-8+\frac{16+8}{6}\]                 \[=-8+\frac{24}{6}\]                 \[=-8+4\] \[=-4\]


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