Question

If a normal is drawn a point ‘P’ of ellipse x²/a² + y²/b² = 1, then Prove that the maximum distance from centre of ellipse will be a – b.

Answer 1

We have to show that maximum value of OM is a – b.

Now equation of normal at point P (a cos θ, b sin θ), ax sec θ – by cosec θ = a² – b²

Perpendicular distance between normal and centre of ellipse, will be minimum.

Here (a² – b²) is constant.

Now, Let y = a² sec² θ + b² cosec² θ

Note : Here, we will prove that (a + b)² is minimum

value of a² sec² θ + b² cosec² θ

dy/dx = 2a² sec² θ tan θ – 2b² cosec² θ cot θ using dy/dx= 0

⇒ 2[a² sec² θ tan θ – b² cosec² θ cot θ] = 0

So, y will be maximum.

Putting tan² θ = b/a in y = a² sec² θ + b² cosec² θ

y = a² sec² θ + b² cosec² θ

= a² (1 + tan² θ) + b² (1 + cot² θ)

= a² [1 + b/a] + b² [1 + a/b]

= a² + b² + 2ab

= (a + b)²

So, y = a² sec² θ + b² cosec² θ has minimum value (a + b)²

From equation (i),

Maximum value of p

Hence, maximum distance between normal and centre of ellipse is (a - b)