• # question_answer The minimum area of a triangle formed by any tangent to the ellipse$\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{81}=1$and the co-ordinate axes is:   [JEE Main Online Paper ( Held On 12 Apirl  2014 ) A) 12                                           B) 18 C) 26                                           D) 36

Let (h, k) be the point on ellipse through which tangent is passing. Equation of tangent at $(h,k)=\frac{xh}{16}+\frac{yk}{81}=1$ at $y=0,x=\frac{16}{h}$ at$x=0,y=\frac{81}{k}$ Area of $AOB=\frac{1}{2}\times \left( \frac{16}{h} \right)\times \left( \frac{81}{k} \right)=\frac{648}{hk}$ ${{A}^{2}}=\frac{{{(648)}^{2}}}{{{h}^{2}}{{k}^{2}}}$                                                       ?(1) (h, k) must satisfy equation of ellipse $=\frac{{{h}^{2}}}{16}+\frac{{{k}^{2}}}{81}=1$ ${{h}^{2}}=\frac{16}{81}+(81-{{k}^{2}})$ Putting value of ${{h}^{2}}$in equation (1) ${{A}^{2}}=\frac{81{{(648)}^{2}}}{16\times {{k}^{2}}(81-{{k}^{2}})}=\frac{\alpha }{81{{k}^{2}}-{{k}^{4}}}$ differentiating w.r. to k $2A{{A}^{'}}=\alpha \left( \frac{-1}{81{{k}^{2}}-{{k}^{4}}} \right)=(162k-4{{k}^{3}})$ $2A{{A}^{'}}=-2A(81k-4{{k}^{3}})\Rightarrow A'=-81k-4{{k}^{3}}$ Put$A'=0$ $\Rightarrow$$162k-4{{k}^{3}}=0,k(162-4{{k}^{2}})=0$ $\Rightarrow$$k=0,k=\pm \frac{9}{\sqrt{2}}$ $A''=-(81-12{{k}^{2}})$ For both value of $k,A''=405>0$ Area will be minimum for $k=\pm \frac{9}{\sqrt{2}}$ ${{h}^{2}}=\frac{16}{81}(81-{{k}^{2}})=8$ $h=\pm 2\sqrt{2}$ Area of triangle $AOB=\frac{648\times \sqrt{2}}{2\sqrt{2}\times 9}=36$sq unit