The locus of the midpoint of the portion of the line $x \cos \alpha+y \sin \alpha=p$ intercepted by the…
The locus of the midpoint of the portion of the line $x \cos \alpha+y \sin \alpha=p$ intercepted by the coordinate axes, where $p$ is a constant, is
- $\frac{1}{x^2}+\frac{1}{y^2}=\frac{3}{p^2}$
- $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$
- $x^2+y^2=2 p^2$
- $\frac{2}{x^2}+\frac{2}{y^2}=\frac{1}{p^2}$
Solution
Given, $x \cos \alpha+y \sin \alpha=p$ ...(i)
Let $P(h, k)$ be the mid point of above line. When $x=0$,
Eq. (i) becomes $y \sin \alpha=p \Rightarrow y=\frac{p}{\sin \alpha}$
When $y=0$, Eq. (i) becomes $x \cos \alpha=p \Rightarrow x=\frac{p}{\cos \alpha}$
$\therefore$ The line meets the coordinate axes at
$A\left(\frac{p}{\cos \alpha}, 0\right), B\left(0, \frac{p}{\sin \alpha}\right)$
Midpoint $P(h, k)=\left(\frac{p}{2 \cos \alpha}, \frac{p}{2 \sin \alpha}\right)$
$\Rightarrow \cos \alpha=\frac{p}{2 h} \text { and } \sin \alpha=\frac{p}{2 k}$
Squaring and adding the above equations,
$\cos ^2 \alpha+\sin ^2 \alpha=\frac{p^2}{4 h^2}+\frac{p^2}{4 k^2} \Rightarrow \frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2}$
$\therefore$ Locus of $P$ is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$.
Asked in: AP EAMCET 2024 (23 May Shift 1)
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