MHT CET · Maths · Parabola
The equation of directrix is to the parabola \(4 x^{2}-4 x-2 y+3=0\) will be
- A \(2 y=1\)
- B \(2 x=1\)
- C \(2 y=3\)
- D \(2 x=3\)
Answer & Solution
Correct Answer
(A) \(2 y=1\)
Step-by-step Solution
Detailed explanation
Given parabola, \(4 x^{2}-4 x-2 y+3=0\) \(\Rightarrow 4\left(x^{2}-x\right)=2 y-3\)
\(\Rightarrow 4\left(x^{2}-x+\frac{1}{4}-\frac{1}{4}\right)=2 y-3\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^{2}=2 y-2\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^{2}=2(y-1)\)
which is of the form \(X^{2}=4 a Y\)
where, \(\quad X=x-\frac{1}{2}\) and \(Y=y-1\)
and \(\quad a=\frac{1}{2}\)
\(\therefore\) Directrix, \(Y+a=0\) \(\Rightarrow \quad y-1+\frac{1}{2}=0\)
\(\Rightarrow \quad y-\frac{1}{2}=0\)
\(\Rightarrow 2 y=1\)
\(\Rightarrow 4\left(x^{2}-x+\frac{1}{4}-\frac{1}{4}\right)=2 y-3\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^{2}=2 y-2\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^{2}=2(y-1)\)
which is of the form \(X^{2}=4 a Y\)
where, \(\quad X=x-\frac{1}{2}\) and \(Y=y-1\)
and \(\quad a=\frac{1}{2}\)
\(\therefore\) Directrix, \(Y+a=0\) \(\Rightarrow \quad y-1+\frac{1}{2}=0\)
\(\Rightarrow \quad y-\frac{1}{2}=0\)
\(\Rightarrow 2 y=1\)
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