Q.203. Ellipse, normal, minima & maxima

Question 203.
( 1 ) Find the minimum value of the segment of a tangent to the ellipse x^2/a^2 + y^2/b^2 = 1       intercepted by the coordinate axis.
 ( 2 ) Find the maximum distance of any normal of the ellipse x^2/a^2 + y^2/b^2 = 1 from its centre.

( 3 ) A normal is drawn to the ellipse x^2 / (a^2+2a+2)^2 + y^2/ (a^2+1)^2 = 1. If maximum radius of the circle centered at the origin and touching the normal is 5, then find possible values of 'a'.

Answer 203.
(1)

The equation of tangent at the 'θ' point to the ellipse is
(x/a) cos θ + (y/b) sin θ = 1
The points of intersection with the axes are
(asecθ, 0) and (0, bcscθ)
The length of segment of tangent between the axes is given by
l^2 = a^2sec^θ + b^2csc^2θ
For l to be minimum, dl/dθ = 0
=> a^2sec^2θ tanθ - b^2csc^2θ cotθ = 0
=> tan^2 θ = b/a
=> l^2 = a^2 (1 + b/a) + b^2(1 + a/b) = (a + b)^2
=> l (minimum) = a + b

(2)
The equation of normal to the ellipse at the 'θ' point is
(asecθ) x - (bcscθ) y = (a^2 - b^2)
Square of perpendicular from origin to the normal
l^2 = (a^2 - b^2)^2 / (a^ sec^2 θ + b^2 csc^2 θ)

For l to be maximum, a^2sec^2θ + b^2 csc^2θ should be minimum => tan^2 θ = b/a (as above)
=> l^2 = (a^2 - b^2)^2 / [a^2(1 + b/a) + b^2(1 + a/b)]
=> l^2 = (a^2 - b^2)^2 / (a + b)^2
=> l = l a - b l

(3)
Using the above result,
± 5 = a^2 + 2a + 2 - a^2 - 1 = 2a + 1
=>  a = 2 or - 3.

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