TS EAMCET · Maths · Straight Lines
A line \(L\) makes intercepts \(a\) and \(b\) on the coordinate axes. The axes are rotated through an angle \(\theta\) in the positive direction, keeping the origin fixed. If the line \(L\) makes intercepts \(p\) and \(q\) on the new coordinate axes, then \(\frac{1}{a^2}+\frac{1}{b^2}=\)
- A \(\frac{1}{p^2 q^2}\)
- B \(\frac{1}{p^2}-\frac{1}{q^2}\)
- C \(\frac{1}{p^2}+\frac{1}{q^2}\)
- D \(\frac{p q}{p^2+q^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{p^2}+\frac{1}{q^2}\)
Step-by-step Solution
Detailed explanation
\( \text { Let the equation of line is } \frac{x}{a}+\frac{y}{b}=1 \text {. } \) Distance from origin \( d=\left|\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right| \Rightarrow \frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2} \) When line is rotated perpendicular distance from origin…
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