JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2010

  • question_answer     The equations of tangents to the ellipse \[9{{x}^{2}}+16{{y}^{2}}=144\]which pass through the point \[(2,3)\]are

    A)  \[y=3,y=x+5\]  

    B)  \[y=3,x=3\]

    C)  \[x=2,\text{ }y=x+5\]   

    D)  \[y=3,\text{ }y=-x+5\]

    Correct Answer: D

    Solution :

                    The equation of ellipse is \[9{{x}^{2}}+16{{y}^{2}}=144\] \[\Rightarrow \]               \[\frac{{{x}^{2}}}{{{4}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]                           ?. (i)      This is of the form \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\therefore \]  \[{{a}^{2}}={{4}^{2}}\]and \[{{b}^{2}}={{3}^{2}}\] The equation of any tangent to the ellipse (i) is \[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\] \[\Rightarrow \]               \[y=mx\pm \sqrt{16{{m}^{2}}+9}\]                   ...(ii) Since, it passes through (2, 3), therefore                 \[3=2m\pm \sqrt{16{{m}^{2}}+9}\] \[\Rightarrow \]               \[{{(3-2m)}^{2}}=(\pm \sqrt{16{{m}^{2}}+9}){{)}^{2}}\] \[\Rightarrow \]               \[9+4{{m}^{2}}-12m=16{{m}^{2}}+9\] \[\Rightarrow \]               \[12{{m}^{2}}+12m=0\] \[\Rightarrow \]               \[{{m}^{2}}+m=0\] \[\Rightarrow \]               \[m(m+1)=0\] \[\Rightarrow \]               \[m=0,-1\] Substituting these values of\[m\]in Eq. (ii), we get \[y=3\] and        \[y=-x+5\]which are the required equation of tangents.

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