A) \[{{a}^{2}}+{{b}^{2}}={{p}^{2}}+{{q}^{2}}\]
B) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}\]
C) \[{{a}^{2}}+{{p}^{2}}={{b}^{2}}+{{q}^{2}}\]
D) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{p}^{2}}}=\frac{1}{{{b}^{2}}}+\frac{1}{{{q}^{2}}}\]
Correct Answer: B
Solution :
Suppose we rotate the coordinate axes in the anti clockwise direction through an angle \[\alpha \]. The equation of the line L with respect to old axes is \[\frac{x}{a}+\frac{y}{b}=1\]. In this question replacing x by \[x\cos \alpha -y\sin \alpha \] and \[y\] by \[x\sin \alpha +y\cos \alpha \], the equation of the line with respect to new axes is \[\frac{x\cos \alpha -y\sin \alpha }{a}+\frac{x\sin \alpha +y\cos \alpha }{b}=1\] Þ \[x\left( \frac{\cos \alpha }{a}+\frac{\sin \alpha }{b} \right)+y\left( \frac{\cos \alpha }{b}-\frac{\sin \alpha }{a} \right)=1\] .....(i) The intercepts made by (i) on the co-ordinate axes are given as p and q. Therefore \[\frac{1}{p}=\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}\]and \[\frac{1}{q}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}\] Squaring and adding, we get \[\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\]. Note: Students should remember this question as a formula.
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