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Question

If the tangent to the ellipse $$x^2+4y^2=16$$ at the point $$P(\phi)$$ is a normal to the circle $$x^2+y^2-8x-4y=0$$ then $$\phi$$ is equal to


A
π2
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B
π4
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C
tan112
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D
π4
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Solution

The correct option is B $$tan^{-1}\frac{1}{2}$$
$$y=mx+\sqrt {16m^2-4}$$
$$c=\pm \sqrt {a^2m^2+b^2}$$
$$\sqrt {16m^2+4}$$
Now it is normal to circle so it passes through centre
$$\therefore (4, 2)$$
$$2=4m \pm \sqrt {16m^2-4}$$
$$1=2m\pm \sqrt {4m^2-1}$$
$$(1-2m)^2=(4m^2-1)$$
$$4m^2-4m+1=4m^2-1$$
$$4m=2, m=\frac {1}{2}$$
$$\phi =tan^{-1}\frac {1}{2}$$

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