If the tangent at a point on the ellipse $\frac{x^2}{27} + \frac{y^2}{3} =1$ meets the coordinate axes at A and B, and O is the origin, them the minimum area (in s units) of the triangle OAB is:
- $\frac{9}{2}$
- $3 \sqrt{3}$
- $9 \sqrt{3}$
- $9$
The Correct Option is D
Solution and Explanation
Equation of tangent to ellipse
\(\frac{x}{\sqrt{27}}+\frac{y}{\sqrt{3}}=1\)
Area bounded by line and co-ordinate axis
\(\frac12\times\)intercept on x-axis \(\times\) intercept on y -axis
\(\Delta=\frac{1}{2}.\frac{\sqrt{27m^2+3}}{m}. {\sqrt{27m^2+3}}{sin}\)
\(\frac12\times\frac{(27m^2+3)}{m}\)
now apply
AM≥GM
\(\frac{27m+\frac3m}{2}\)≥\(\sqrt{27m\times\frac3m}\) ≥ \(9\)
\(\Delta\)
\(\Delta_{min}=9\)
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