Question

Angle between the tangents drawn from point (4 5) to the ellipse x2/16+y2/25=1 is

Answer 1

$\theta = tan^{ - 1} ( \frac{ - 5}{4} )$

Step-by-step explanation:

$Given, \: \: \: p(4,5) \: \: and \: \: \frac{ {x}^{2} }{16} + \frac{ {y}^{2} }{25} = 1 \\ we \: \: know \: \: \boxed{m = tan \theta \: = \frac{dy}{dx} } \\ Now, \: \: Differentiating \: \: \frac{ {x}^{2} }{16} + \frac{ {y}^{2} }{25} = 1 \: \: with \: \: respect \: \: to \: \: x, \\ \frac{2x}{16} + \frac{2y}{25} \frac{dy}{dx} = 0 \\ \frac{2y}{25} \frac{dy}{dx} = - \frac{2x}{16} \\ \frac{dy}{dx} = \frac{ -x}{16} \times \frac{25}{y} \\ tan \theta = \frac{ - 25x}{16y} \\ putting \: \: value \: \: p(x, y) = (4, 5) \\ tan \theta = \frac{ - 25 \times 4}{16 \times 5} = \frac{ - 5}{4} \\ \theta = tan^{ - 1} ( \frac{ - 5}{4} )$