KCET · Maths · Parabola
The length of the latusrectum of \(3 x^{2}-4 y+6 x-3=0\) is
- A \(\frac{3}{4}\)
- B \(\frac{4}{3}\)
- C 2
- D 3
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
Given equation of conic
\[
3 x^{2}-4 y+6 x-3=0
\]
\(\begin{array}{lc}\Rightarrow & 3 x^{2}+6 x-3=4 y \\ \Rightarrow & 3\left(x^{2}+2 x-1\right)=4 y \\ \Rightarrow & 3\left(x^{2}+2 x+1-2\right)=4 y \\ \Rightarrow & 3(x+1)^{2}-6=4 y \\ \Rightarrow & 3(x+1)^{2}=4 y+6 \\ \text { Let } & (x+1)^{2}=\frac{4}{3}(y+3 / 2) \\ & X^{2}=\frac{4}{3} Y\end{array}\)
\(X^{2}=\frac{4}{3} Y\)
where \(X=x+1\) and \(Y=y+3 / 2\)
and \(\quad 4 b=4 / 3 \Rightarrow b=1 / 3\)
So, now the length of latusrectum is \(4 / 3\).
\[
3 x^{2}-4 y+6 x-3=0
\]
\(\begin{array}{lc}\Rightarrow & 3 x^{2}+6 x-3=4 y \\ \Rightarrow & 3\left(x^{2}+2 x-1\right)=4 y \\ \Rightarrow & 3\left(x^{2}+2 x+1-2\right)=4 y \\ \Rightarrow & 3(x+1)^{2}-6=4 y \\ \Rightarrow & 3(x+1)^{2}=4 y+6 \\ \text { Let } & (x+1)^{2}=\frac{4}{3}(y+3 / 2) \\ & X^{2}=\frac{4}{3} Y\end{array}\)
\(X^{2}=\frac{4}{3} Y\)
where \(X=x+1\) and \(Y=y+3 / 2\)
and \(\quad 4 b=4 / 3 \Rightarrow b=1 / 3\)
So, now the length of latusrectum is \(4 / 3\).
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