Question

Find the equation of pair of tangents to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ from (5,4)

• A. $4x+5y-xy=20$
• B. $4x+5y-xy=21$
• C. $4x+5y+xy=40$
• D. $5x+4y+xy=2$

Answer 1

The correct option is A

$4x+5y-xy=20$

The equation of a pair of tangent to any second degree curve

$S=a{x}^2+2hxy+b{y}^2+29x+2fy+c=0$

is given by $S{S}_1=T^2$,

where $S_1=a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2+29x_1+2f{y}_1+c=0$

T is obtained by replacing x by $\frac{x+x_1}{2}Yby\frac{y+y_1}{2},x^{2}byx{x}_1,y^{2}byyy,andxyby\frac{x{y}_1+y{x}_1}{2}$.

In our case, $(x_1,y_1)=(5,4)and‘\frac{x^2}{25}+\frac{y^2}{16}=1$ is the equation of curve

$\Rightarrow S=\frac{x^2}{25}+\frac{y^2}{16}-1$ is the equation of curve

$\Rightarrow S_1=\frac{5^2}{25}+\frac{4^2}{16}-1=1$

$T=\frac{5x}{25}+\frac{4y}{16}-1=\frac{x}{5}+\frac{y}{4}-1$

$S{S}_1=T^2$

$\Rightarrow (\frac{x^2}{25}+\frac{y^2}{16}-1)\times 1={(\frac{x}{5}+\frac{y}{4}-1)}^2$

$\Rightarrow \frac{x^2}{25}+\frac{y^2}{16}-1=\frac{x^2}{25}+\frac{y^2}{16}+1-\frac{2x}{5}-\frac{y}{2}+\frac{xy}{10}$

$\Rightarrow \frac{2x}{5}+\frac{y}{2}-\frac{xy}{10}=2or4x+5y-xy=20$