A) \[x=\frac{{{B}^{2}}+{{A}^{2}}-C}{2A}\]
B) \[x=\frac{{{B}^{2}}-{{A}^{2}}+C}{2A}\]
C) \[x=\frac{{{B}^{2}}-{{A}^{2}}-C}{2A}\]
D) \[x=\frac{{{A}^{2}}-{{B}^{2}}-C}{2A}\]
Correct Answer: B
Solution :
\[{{(y+B)}^{2}}=-2Ax-C+{{B}^{2}}=-2A\left( x+\frac{C}{2A}-\frac{{{B}^{2}}}{2A} \right)\] Equation of latus rectum \[x+\lambda =0\] Vertex \[=\left( \frac{-C+{{B}^{2}}}{2A},B \right)\], focus \[\equiv \left( \frac{-C+{{B}^{2}}}{2A}-\frac{A}{2},B \right)\] Equation of L.R. is \[x=\frac{-C+{{B}^{2}}}{2A}-\frac{A}{2}=\frac{{{B}^{2}}-{{A}^{2}}-C}{2A}\].
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