WBJEE · Maths · Straight Lines
If the vertex of the conic \(y^{2}-4 y=4 x-4 a\) always lies between the straight lines \({x}+y=3\) and \(2 x+2 y-1=0,\) then
- A \(2 < a < 4\)
- B \(-\frac{1}{2} < a < 2\)
- C \(0 < a < 2\)
- D \(-\frac{1}{2} < a < \frac{3}{2}\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{2} < a < 2\)
Step-by-step Solution
Detailed explanation
Equation of the vertex of conic \[ y^{2}-4 y=4 x-4 a \] \(\Rightarrow \quad y^{2}-4 y+4=4 x-4 a+4\) \(\Rightarrow \quad(y-2)^{2}=4[x-(a-1)]\) \(\therefore\) Vertex \(=(a-12)\) From figure, clearly. \(-\frac{3}{2} < a-1 < 1\) \(\therefore\) \(-\frac{1}{2} < a < 2\)
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