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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

```
9 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....
```
9 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 2tFor minimum area sin2t should be maximum that is 1,So, Area= ab
```
3 years ago
```							Equation of tangent to given ellipse isy=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
```
3 months ago
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