Question

If $a^2+b^2=1$, and $x^2+y^2=1$, show that $ax+by<1$.

Answer 1

Correct relation is $ax+by\le 1$

Solving this question using parametric equation of circle.

Since $a^2+b^2=1$ and $x^2+y^2=1$ both are equations of circle so let say $a=\cos \theta$ and $x=\cos \alpha$ then $b=\sin \theta$ and $y=\sin \alpha$

so $ax+by=\left\{\cos \theta \right\}\left\{\cos \alpha \right\}+\left\{\sin \theta \right\}\left\{\sin \alpha \right\}$ $=\cos \{\theta -\alpha \}$

And as we know $\cos \{\theta -\alpha \}\le 1$ so $ax+by\le 1$.