Mathematics · 2014
JEE Main · 9 April 2014, Shift 1 · Q79
Let a and b be any two numbers satisfying 1/(a^2)+1/(b^2)=1/4. Then, the foot of perpendicular from the origin on the variable line, x/a+y/b=1, lies…
Let a and b be any two numbers satisfying $\displaystyle \frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}=\frac{1}{4}$. Then, the foot of perpendicular from the origin on the variable line, $\displaystyle \frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}=1$, lies on :
Official answer
(3)
a circle of radius $\displaystyle =2$
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JEE Main 2014 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.