Find the equations of the tangents to the circle $\;$ $2x^2 + 2y^2 = 5$ $\;$ which are perpendicular to $\;$ $y = 2x$.
Given circle: $\;\;$ $2x^2 + 2y^2 = 5$
i.e. $\;$ $x^2 + y^2 = \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 = a^2$ $\;$ gives
$a^2 = \dfrac{5}{2}$ $\implies$ $a = \pm \sqrt{\dfrac{5}{2}}$
Given line: $\;\;$ $y = 2x$ $\;\;\; \cdots \; (2)$
Slope of the given line $= m_1 = 2$
Since the required tangents are perpendicular to $(2)$,
$\therefore \;$ slope of tangents $= m = \dfrac{-1}{m_1} = \dfrac{-1}{2}$
Equations of tangents are
$y = mx \pm a \sqrt{1 + m^2}$
i.e. $\;$ $y = \dfrac{-1}{2}x \pm \sqrt{\dfrac{5}{2}} \times \sqrt{1 + \dfrac{1}{4}}$
i.e. $\;$ $y = \dfrac{-1}{2}x \pm \sqrt{\dfrac{5}{2}} \times \dfrac{\sqrt{5}}{2}$
i.e. $\;$ $y = \dfrac{-1}{2} x \pm \dfrac{5}{2 \sqrt{2}}$
i.e. $\;$ $2 \sqrt{2} y = - \sqrt{2} x \pm 5$
$\therefore \;$ The equations of the required tangents are
$\sqrt{2} x + 2 \sqrt{2} y + 5 = 0$ $\;$ and $\;$ $\sqrt{2} x + 2 \sqrt{2} y - 5 = 0$