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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
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....
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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