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If the pair of lines ax^{2}+2hxy+by^{2}+2gx+2fy+c=0    intersect on the y-axis then    

  • Option 1)

    2f\! gh=bg^{2}+ch^{2}\;

  • Option 2)

    \; \; bg^{2}\neq ch^{2}\;

  • Option 3)

    \; \; abc=2f\! gh\; \;

  • Option 4)

    none of these

 

Answers (2)

As we learned  @19610

 

x=0:by^2+2fy+c=0

Since, equal roots,

\Rightarrow D=0

\Rightarrow 4f^2-4bc=0

\Rightarrow f^2=bc............................(i)

Now,

bg^2+ch^2-2fgh= bg^2+ch^2-2\sqrt{bc}\: gh....................from(i)

Also,

\Delta = 0

\Rightarrow abc+2fgh-af^2-bg^2-ch^2=0

\Rightarrow2\sqrt{bc}\: gh-bg^2-ch^2=0................\left ( \because f^2=bc \right )

\Rightarrow\left ( \sqrt{ch}-\sqrt{bg} \right )= 0

\Rightarrow bg^2+ch^2-2fgh= 0= \left ( \sqrt{b}g-\sqrt{c}h \right )^{2}

 

 

 


Option 1)

2f\! gh=bg^{2}+ch^{2}\;

Option 2)

\; \; bg^{2}\neq ch^{2}\;

Option 3)

\; \; abc=2f\! gh\; \;

Option 4)

none of these

Posted by

Vakul

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Posted by

Samhitha

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