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Botany-

General

Easy

Question

If the tangent at Point P to the ellipse 16x² + 11y² = 256 is also the tangent to the circle x² + y² - 2x = 15, then the eccentric angle of point P is

$\pm \frac{π}{2}$
$\pm \frac{π}{4}$
$\pm \frac{π}{3}$
$\pm \frac{π}{6}$

The correct answer is: $plus-or-minus fraction numerator pi over denominator 3 end fraction$

The equation of tangent at point to the ellipse
16x² + 11y² = 256 is
16x (4cosq) + 11y = 256
4xcosq + y sinq = 16
This touches the circle
(x + 1)² + y² = 16
So, = 4
Þ (cosq - 4)² = 11 + 5cos²q
4cos²q + 8cosq - 5 = 0
\ cosq =
\ q = ±

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