MHT CET · Maths · Straight Lines
The locus of a point of intersection of two lines \(x \sqrt{3}-y=k \sqrt{3}\) and
\(\sqrt{3} k x+k y=\sqrt{3}, k \in R\), describes
- A a parabola
- B a hyperbola
- C an ellipse
- D a pair of lines
Answer & Solution
Correct Answer
(B) a hyperbola
Step-by-step Solution
Detailed explanation
We have \(\begin{aligned} \quad x \sqrt{3}-y &=k \sqrt{3} ...(1)\\ \sqrt{3} k x+k y &=\sqrt{3} ...(2) \end{aligned}\)
We will equate value of \(k\) from eq. (1) and (2)
\( \frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3} x+y} \)
\( \therefore (x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3}) \)
\( \therefore 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3} \text { i.e. } \frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}\)\(=1, \text { which is equation of hyperbola. }\)
We will equate value of \(k\) from eq. (1) and (2)
\( \frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3} x+y} \)
\( \therefore (x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3}) \)
\( \therefore 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3} \text { i.e. } \frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}\)\(=1, \text { which is equation of hyperbola. }\)
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