Question

The equations of the tangents to the ellipse $9x^2+16y^2=144$ which passes through the point $(2,3)$ is

• A. $y=3,x+y=5$
• B. $y=-3,x-y=5$
• C. $y=4,x+y=3$
• D. $y=4,x-y=3$

Answer 1

Answer: (A)

For ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$
the combined equation of tangent from $(2,3)$ i.e.,
$T^2=S{S}_1\text{ i.e., }{\left(\frac{x}{8}+\frac{y}{3}-1\right)}^2=\frac{1}{4}\left(\frac{x^2}{16}-\frac{y^2}{9}-1\right)$
$\Rightarrow 9x^2+64y^2+576-144x-384y+48xy$
$=9x^2+16y^2-144$
or $48y^2+48xy-384y-144x+720=0$
$\Rightarrow$ i.e., $y^2+xy-3x-8y+15=0$
$\Rightarrow (y-3)(x+y-5)=0$
$\Rightarrow y=3$ and $x+y=5$ are the tangent