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MHT CET · Maths · Straight Lines

\(\mathrm{p}\) is the length of perpendicular from the origin to the line whose intercepts on the axes are a and \(\mathrm{b}\) respectively, then \(\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}\) equals

  1. A \(\mathrm{p}^2\)
  2. B \(\frac{2}{\mathrm{p}^2}\)
  3. C \(\frac{1}{\mathrm{p}^2}\)
  4. D \(\frac{1}{2 p^2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{\mathrm{p}^2}\)

Step-by-step Solution

Detailed explanation

Let the equation of the line be \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}=1\)
According to the given condition,
\(
\begin{aligned}
& \mathrm{p}=\left|\frac{\mathrm{ab}}{\sqrt{\mathrm{a}^2+b^2}}\right| \\
& \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2 \mathrm{~b}^2}=\frac{1}{\mathrm{p}^2} \\
& \Rightarrow \frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}=\frac{1}{\mathrm{p}^2}
\end{aligned}
\)
From MHT CET
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