A) \[\frac{x}{a}+\frac{y}{b}=1\]
B) \[\frac{x}{a}+\frac{y}{b}+1=0\]
C) \[\frac{x}{a}-\frac{y}{b}=1\]
D) \[\frac{x}{a}-\frac{y}{b}+1=0\]
Correct Answer: A
Solution :
From formula of tangent at a point, \[x\left( \frac{a{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \right)+y\left( \frac{{{a}^{2}}b}{{{a}^{2}}+{{b}^{2}}} \right)=\frac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\Rightarrow \frac{x}{a}+\frac{y}{b}=1\].You need to login to perform this action.
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