If the tangent to the ellipse x2-4 y2-16 at P- is normal to circle x2-y2-8 x-4 y-0- Then - equals toA-B-C- 0D- -4

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If the tangen...

Question

If the tangent to the ellipse $x^{2} + 4 y^{2} = 16$ at $P (θ)$ is normal to circle $x^{2} + y^{2} - 8 x - 4 y = 0$ . Then $θ$ equals to

$\frac{π}{2}$

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$\frac{π}{4}$

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$- \frac{π}{4}$

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x^{2} + 4 y^{2} = 16

^{'} ϕ^{'}

x^{2} + y^{2} - 8 x - 4 y = 0

ϕ

x^{2} + 4 y^{2} = 16

P (ϕ)

x^{2} + y^{2} - 8 x - 4 y = 0,

ϕ

A. If the tangent to the ellipse $x^{2} + 4 y^{2} = 16$ at the point $P (ϕ)$ is a normal to the circle $x^{2} + y^{2} - 8 x - 4 y = 0$ , then $\frac{ϕ}{2}$ may be	1. $0$
B. The eccentric angle(s) of a point on the ellipse $x^{2} + 3 y^{2} = 6$ at a distance $2$ units from the centre of the ellipse is/are	2. ${cos}^{- 1} (- \frac{2}{3})$
C. The eccentric angle of intersection of the ellipse $x^{2} + 4 y^{2} = 4$ and the parabola $x^{2} + 1 = y$ is	3. $\frac{π}{4}$
D. If the normal at the point $P (θ)$ to the ellipse $\frac{x^{2}}{14} + \frac{y^{2}}{5} = 1$ intersects it again at the point $Q (2 θ)$ , then $θ$ is	4. $\frac{5 π}{4}$

P (θ)

16 x^{2} + 11 y^{2} = 256

x^{2} + y^{2} - 2 x = 15,

θ

x^{2} + 4 y^{2} = 16

O

x^{2} = 8 (y - 6)

θ

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