MHT CET · Maths · Straight Lines
If \(\mathrm{p}\) is the length of the perpendicular from origin to the whose intercepts on the axes are \(a\) and \(b\), then \(\frac{1}{a^2}+\frac{1}{b^2}=\)
- A \(\mathrm{p}^2\)
- B \(\frac{1}{2 p^2}\)
- C \(2 \mathrm{p}^2\)
- D \(\frac{1}{\mathrm{p}^2}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{\mathrm{p}^2}\)
Step-by-step Solution
Detailed explanation
Refer image

Equation of given line is
Distance of line (1) from origin is
\(
\begin{aligned}
& \frac{|-a b|}{\sqrt{a^2+b^2}}=p \quad \Rightarrow a^2+b^2=\frac{a^2 b^2}{p^2} \\
& \therefore \frac{a^2+b^2}{a^2 b^2}=\frac{1}{p^2} \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}
\end{aligned}
\)

Equation of given line is

Distance of line (1) from origin is
\(
\begin{aligned}
& \frac{|-a b|}{\sqrt{a^2+b^2}}=p \quad \Rightarrow a^2+b^2=\frac{a^2 b^2}{p^2} \\
& \therefore \frac{a^2+b^2}{a^2 b^2}=\frac{1}{p^2} \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}
\end{aligned}
\)
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