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# find the min area of the triangle formed by the tangents to the ellipse x^2/a^2 +y^2/b^2 =1and the coordinate axis

## 3 Answers

10 years ago

eq of tangent to ellipse is given by

y = mx +(-) (a2m2+b2)1/2

it cuts the coordinate axis at  [ 0, (a2m2+b2)1/2  ]     &      [ (a2m2+b2)1/2/m , 0 ]

area formed by these points & origin is

A = (1/2) [ (a2m2+b2)/m ]

now using maxima minima concept

diffenentiating this eq wrt m & after putting it  to 0 we get

m2a2 - b2 = 0

m = +(-) (b/a)

so , area can be

A = (2b2)a/2b = ab                ( at m = b/a)

this is the minimum area of the triangle formed by tangent and coordinate axis....

4 years ago
At a parametric point (a cost , b sint)
Equation of the tangent is bx cost + ay sint = ab,
Therefore, X intercept is a/cost and Y intercept is b/sint,
Hence,
Area  = (ab)/2*sint*cost
= ab/sin 2t
For minimum area sin2t should be maximum that is 1,
So, Area= ab
10 months ago

Equation of tangent to given ellipse is

y=mx+a2m2+b2−−−−−−−−√

It meets axes at (−a2m2+b2−−−−−−−√m,0) and (0,a2m2+b2−−−−−−−−√)

∴ Area of triangle

=12∣∣a2m2+b2m∣∣

=12∣∣∣a2m+b2m∣∣≥ab

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