Straight Line Question 139
Question: If for a variable line $ \frac{x}{a}+\frac{y}{b}=1 $ , the condition $ \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}} $ (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is [RPET 1999]
Options:
A) $ x^{2}+y^{2}=c^{2}/2 $
B) $ x^{2}+y^{2}=2c^{2} $
C) $ x^{2}+y^{2}=c^{2} $
D) $ x^{2}-y^{2}=c^{2} $
Correct Answer: CShow Answer
Answer:
Solution:
$ \therefore m $ of perpendicular $ . =\frac{a}{b} ] $
Þ $ by-ax=0 $
Þ $ \frac{x}{b}-\frac{y}{a}=0 $ Now, the locus of foot of perpendicular is the intersection point of line $ \frac{x}{a}+\frac{y}{b}=1 $ …..(i) and $ \frac{x}{b}-\frac{y}{a}=0 $ ……(ii) To find locus, squaring and adding (i) and (ii) $ {{( \frac{x}{a}+\frac{y}{b} )}^{2}}+{{( \frac{x}{b}-\frac{y}{a} )}^{2}}=1 $
Þ $ x^{2}( \frac{1}{a^{2}}+\frac{1}{b^{2}} )+y^{2}( \frac{1}{a^{2}}+\frac{1}{b^{2}} )=1 $
Þ $ x^{2}( \frac{1}{c^{2}} )+y^{2}( \frac{1}{c^{2}} )=1 $ , $ [ \because \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}} ] $
Þ $ x^{2}+y^{2}=c^{2} $ .
BETA
