Q.189. Normal to an ellipse.

Question 189.
What is the maximum distance of a nomal to the ellipse
b^2 x^2 +a^2 y^2 = a^2 b^2 from its centre?

Answer 189.
b^2 x^2 + a^2 y^2 = a^2 b^2
=> 2b^2 x + 2a^2 y dy/dx = 0
=> dy/dx = - (b^2 x) / (a^2 y)
 => slope of the normal passing through any point P = (acosθ, bsinθ) of the ellipse is
m = (a^2 * bsinθ) / (b^2 * acosθ) = (asinθ) / (bcosθ)

Equation of the normal at P is
y - bsinθ = [(asinθ) / (bcosθ)] (x - acosθ)
=> (bcosθ) y - b^2 sinθ cosθ = (asinθ) x - a^2 sinθ cosθ
=> (asinθ) x - (bcosθ) y + (a^2 - b^2) sinθ cosθ = 0
=> [asecθ / (a^2 – b^2)] x + [bcosecθ / (a^2 – b^2)] y + 1 = 0

Perpendicular distance, p, from the center (0, 0) to the above normal is given by
p^2 = (a^2 – b^2)^2 / (a^2 sec^2 θ + b^2 cosec^2 θ)

For p to be maximum, p^2 is maximum
=> dp^2/dθ = 0

dp^2/dθ = 0
=> 2a^2 sec^2 θ tanθ – 2b^2 cosec^2 θ cotθ = 0
=> a^2 sin^4 θ – b^2 cos^4 θ = 0
=> tan^2 θ = (b/a)
=> sec^2 θ = 1 + b/a = (a + b) / a
and cosec^2 θ = 1 + a/b = (a + b) / b

=> p^2
= (a^2 - b^2)^2 / [a(a +b) + b(a + b)]
= (a - b)^2 * (a + b)^2 / (a + b)^2
= (a - b)^2
=> p = l a - b l.

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