• # 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)  $y=3,\text{ }y=-x+5$

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.