Question

if p is the length of perpendicular drawn from the origin to any normal to the ellipse x^2/25 + y^2 / 16 = 1 then the maximum value of p is

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This also is helpful.
https://math.stackexchange.com/questions/305534/max-distance-of-normal-to-ellipse-from-origin

Answer 1

We know that the equation of an normal (in parametric form), at point (a cosθ,b sinθ) is,

ax(secθ) - by(cosecθ)=a²-b².

Now consider a parametric point 'P' on the ellipse [ (x²/5²)+(y²/4²)=1] which is:                       P(5 cosθ,4 sinθ) .

∴The equation of the normal at this point becomes ,
x(5 secθ)-y(4 cosecθ)=5²-4²=9,i.e

x(5 secθ)-y(4 cosecθ)-9=0

Now we can calculate the perpendicular distance 'p' from the origin as follows,

p=|(0)(5 secθ)-(0)(4 cosecθ)-9|÷[√(5²sec²θ+4²cosec²θ)]
p=|-9|÷[√(25 sec²θ+16 cosec²θ)]
For maximum value of 'p',we have to minimize the denominator (25 sec²θ+16 cosec²θ)

Let ƒ(θ)=25 sec²θ+16 cosec²θ
On equating d(ƒ(θ))÷dθ=0,
We see that f(θ) will attain the minimum value at tan²θ=4÷5
∴secθ=3/√5  & cosecθ=3/2
On solving we get,
p=[9÷√(81)]=1