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 = 61

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
$S_1=a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2+29x_1+2f{y}_1+c$
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
$\Rightarrow S_1=\frac{5^2}{25}+\frac{4^2}{16}-1=1T=\frac{5x}{25}+\frac{4y}{16}-1=\frac{x}{5}+\frac{y}{4}-1S{S}_1=T^2\Rightarrow (\frac{x^2}{25}+\frac{y^2}{16}-1)\times 1={(\frac{x}{5}+\frac{y}{4}-1)}^2\Rightarrow \frac{x^2}{25}+\frac{y^2}{16}-1=\frac{x^2}{25}+\frac{y^2}{16}+1-\frac{2x}{5}-\frac{y}{2}+\frac{xy}{10}\Rightarrow \frac{2x}{5}+\frac{y}{2}-\frac{xy}{10}=2or4x+5y-xy=20$