Question

The common tangents to the circle x2+y2=2 and the parabola y2=8x

Answer 1

$the \: \: common \: \: tangents \: \: at \: \: x = - 4$

Step-by-step explanation:

$given \\ circle \: \: {x}^{2} + {y}^{2} = 2 \: \: and \: \: {y}^{2} = 8x \\ we \: \: know \: \: that \\ \boxed{ tan \theta = \frac{dy}{dx} } \\ now \: \: for \: \: tangent \: \: of \: circle \\ {x}^{2} + {y}^{2} = 2 \\ Differentiating \: \: with \: \: respect \: \: to \: \: x, \\ 2x + 2y \frac{dy}{dx} = 0 \\ x + y\frac{dy}{dx} = 0 \\ \frac{dy}{dx} = \frac{ - x}{y} \\ again \: \: for \: \: tangent \: \: of \: \: parabola \\ {y}^{2} = 8x \\ Differentiating \: \: with \: \: respect \: \: to \: \: x, \\ 2y \frac{dy}{dx} = 8 \\ \frac{dy}{dx} = \frac{4}{y} \\ now \: \: A/Q, \\ \frac{ - x}{y} = \frac{4}{y} \\ x = - 4 \\ the \: \: common \: \: tangents \: \: at \: \: x = - 4$