Explanation
Number of possible tangents to the curve $$y = \cos \left( {x + y} \right), - 3\pi \leqslant x \leqslant 3\pi $$, that are parallel to the line $$x + 2y = 0$$, is
We have,
$$ \sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{y}{b}}=2\,\,.......\,\,\left( 1 \right) $$
$$ \dfrac{\sqrt{x}}{\sqrt{a}}+\dfrac{\sqrt{y}}{\sqrt{b}}=2 $$
On differentiation and we get,
$$ \dfrac{1}{2\sqrt{x}\sqrt{a}}+\dfrac{1}{2\sqrt{y}\sqrt{b}}\dfrac{dy}{dx}=0 $$
$$ \dfrac{1}{2\sqrt{y}\sqrt{b}}\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{x}\sqrt{a}} $$
$$ \dfrac{dy}{dx}=-\dfrac{\sqrt{y}\sqrt{b}}{\sqrt{x}\sqrt{a}} $$
At the point $$\left( a,\,b \right)$$ and we get,
$$ \dfrac{dy}{dx}=-\dfrac{\sqrt{b}\sqrt{b}}{\sqrt{a}\sqrt{a}} $$
$$ \dfrac{dy}{dx}=-\dfrac{b}{a} $$
Equation of tangent is
$$ y-{{y}_{1}}=\dfrac{dy}{dx}\left( x-{{x}_{1}} \right) $$
$$ y-b=-\dfrac{b}{a}\left( x-a \right) $$
$$ ay-ab=-bx+ab $$
$$ ay+bx=2ab $$
$$ \dfrac{ay+bx}{ab}=2 $$
$$ \dfrac{y}{b}+\dfrac{x}{a}=2 $$
$$ \dfrac{x}{a}+\dfrac{y}{b}=2 $$
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