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Q.
The equations of the tangents to the ellipse $9 x^{2}+16 y^{2}=144$ which passes through the point $(2,3)$ is
Conic Sections
Solution:
For ellipse $\frac{ x ^{2}}{16}+\frac{ y ^{2}}{9}=1$
the combined equation of tangent from $(2,3)$ i.e.,
$ T ^{2}= SS _{1} \text { i.e., }\left(\frac{ x }{8}+\frac{ y }{3}-1\right)^{2}=\frac{1}{4}\left(\frac{ x ^{2}}{16}-\frac{ y ^{2}}{9}-1\right)$
$\Rightarrow 9 x ^{2}+64 y ^{2}+576-144 x -384 y +48 xy$
$ =9 x ^{2}+16 y ^{2}-144$
or $ 48 y ^{2}+48 xy -384 y -144 x +720=0$
$\Rightarrow $ i.e., $ y ^{2}+ xy -3 x -8 y +15=0$
$\Rightarrow ( y -3)( x + y -5)=0 $
$\Rightarrow y =3$ and $ x + y =5$ are the tangent