WBJEE · Maths · Parabola
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
We have, \(y^{2}-4 y=4 x-4 a\) \(\Rightarrow \quad(y-2)^{2}-4=4 x-4 a\) \(\Rightarrow \quad(y-2)^{2}=4 x-4 a+4\) \(\Rightarrow \quad(y-2)^{2}=4|x-(a-1)|\) Hence, vertex is \((a-1,2)\) Since, vertex lies between the lines \(x+y=3\) and \(2 x+2 y-1=0,\) then…
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