KCET · Maths · Hyperbola
Locus of a point which moves such that its distance from the \(x\)-axis is twice its distance from the line \(x-y=0\) is
- A \(x^{2}+4 x y-y^{2}=0\)
- B \(2 x^{2}-4 x y+y^{2}=0\)
- C \(x^{2}-4 x y+y^{2}=0\)
- D \(x^{2}-4 x y-y^{2}=0\)
Answer & Solution
Correct Answer
(B) \(2 x^{2}-4 x y+y^{2}=0\)
Step-by-step Solution
Detailed explanation
\(P_{1}=\) length of perpendicular from \(P\) to \(x\)-axis.
\(P_{2}=\) length of perpendicular from \(P\) to \(y=x\) line.
\[
\begin{aligned}
P_{1} &=2 P_{2} \\
|k| &=2 \cdot \frac{|h-k|}{\sqrt{2}}
\end{aligned}
\]
Squaring on both sides,
\[
k^{2}=2(h-k)^{2}
\]

\[
\begin{array}{r}
k^{2}=2 h^{2}+2 k^{2}-4 h k \\
\Rightarrow \quad 2 h^{2}-4 h k+k^{2}=0
\end{array}
\]
So, the locus of a point \(P\) is,
\[
2 x^{2}-4 x y+y^{2}=0
\]
\(P_{2}=\) length of perpendicular from \(P\) to \(y=x\) line.
\[
\begin{aligned}
P_{1} &=2 P_{2} \\
|k| &=2 \cdot \frac{|h-k|}{\sqrt{2}}
\end{aligned}
\]
Squaring on both sides,
\[
k^{2}=2(h-k)^{2}
\]

\[
\begin{array}{r}
k^{2}=2 h^{2}+2 k^{2}-4 h k \\
\Rightarrow \quad 2 h^{2}-4 h k+k^{2}=0
\end{array}
\]
So, the locus of a point \(P\) is,
\[
2 x^{2}-4 x y+y^{2}=0
\]
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