Q113 / 180 · COMEDK 2018 · Mathematics
The locus of the mid-point of the intercept of the line $x \cos \alpha+y \sin \alpha=p$ between the coordinate axes is
- Correct answerA$x^{-2}+y^{-2}=4 p^{-2}$
- B$x^{-2}+y^{-2}=p^{-2}$
- C$x^{2}+y^{2}=4 p^{-2}$
- D$x^{2}+y^{2}=p^{2}$
Solution
Given, line segment is $x \cos \alpha+y \sin \alpha=p$
Therefore, it cuts the $X$-axis and $Y$-axis at points $(p \sec \alpha, 0)$ and $(0, p \operatorname{cosec} \alpha)$.
Hence, the coordinate of the mid point are $\left(\frac{p \sec \alpha}{2}, \frac{p \operatorname{cosec} \alpha}{2}\right)$
Therefore, $x=\frac{p \sec \alpha}{2}$ and $y=\frac{p \operatorname{cosec} \alpha}{2}$
or $\cos \alpha=\frac{p}{2 x}$ and $\sin \alpha=\frac{p}{2 y}$
$\because \quad \sin ^{2} \alpha+\cos ^{2} \alpha=1$
$\left(\frac{p}{2 x}\right)^{2}+\left(\frac{p}{2 y}\right)^{2}=1$
$\Rightarrow \frac{p^{2}}{4 x^{2}}+\frac{p^{2}}{4 y^{2}}=1 \Rightarrow \frac{1}{4 x^{2}}+\frac{1}{4 y^{2}}=\frac{1}{p^{2}}$
$\Rightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{p^{2}} \Rightarrow x^{-2}+y^{-2}=4 p^{-2}$
Hence, the correct answer is option: 'A'.