If the tangent to the ellipse x2-4y2-16 at the point P- is a normal to the circle x2-y2-8x-4y-0 then - is equal to

The correct option is B tan^ 11/2 y=mx+√(16m^2 4) c=±√(a^2m^2+b^2) √(16m^2+4) Now it is normal to circle so it passes through centre ∴ ...

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Standard XII

Mathematics

Point Form of Tangent: Ellipse

If the tangen...

Question

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 $ϕ$ is equal to

\frac{π}{2}

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\frac{π}{4}

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t a n^{- 1} \frac{1}{2}

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\frac{- π}{4}

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Solution

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

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If the tangent to the ellipse x2+4y2=16 at the point P(ϕ) is a normal to the circle x2+y2−8x−4y=0 then ϕ is equal to

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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 $ϕ$ is equal to