Question

Find the locus of the middle point of the portion of the line $x\cos \left(\alpha \right)+y\sin \left(\alpha \right)=p$ which is intercepted between the coordinate axes is, given that $p$ remains constant.

• A. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{p^2}$
• B. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{2p^2}$
• C. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$
• D. none of these

Answer 1

The correct option is C

$\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$

Explanation for the correct option:

Finding the locus of the middle point of a given line:

Given line is:$x\cos \left(\alpha \right)+y\sin \left(\alpha \right)=p$

At the point $A,y=0,\Rightarrow x=\frac{p}{\sin \left(\alpha \right)}$

$\Rightarrow A=\left(\frac{p}{\cos \left(\alpha \right)},0\right)$

At the point $B,x=0,\Rightarrow y=\frac{p}{\sin \left(\alpha \right)}$

$\Rightarrow B=\left(0,\frac{p}{\sin \left(\alpha \right)}\right)$

The midpoint of $AB$ is $P\left(h,k\right)$.

$$\begin{gathered} \Rightarrow h=\frac{p}{2\cos \left(\alpha \right)},k=\frac{p}{2\sin \left(\alpha \right)} \\ \Rightarrow \frac{p}{2h}=\cos \left(\alpha \right) \\ \Rightarrow \frac{p}{2k}=\sin \left(\alpha \right) \end{gathered}$$

Squaring and adding, we get

$$\begin{gathered} \Rightarrow \frac{p^2}{4h^2}+\frac{p^2}{4k^2}={\sin }^2\left(\alpha \right)+{\cos }^2\left(\alpha \right) \\ \Rightarrow \frac{p^2}{4h^2}+\frac{p^2}{4k^2}=1 \\ \Rightarrow \frac{1}{4x^2}+\frac{1}{4y^2}=\frac{1}{p^2} \\ \Rightarrow \frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2} \end{gathered}$$

Therefore, the required locus is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$ .

Hence, option $C$ is the correct answer.