Question

Find equation of pair of tangents drawn from point $(6,4)$ to the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$

• A. $(\frac{x^2}{25}+\frac{y^2}{9}-1)={(\frac{6x}{25}+\frac{4y}{9}-1)}^2$
• B. $(\frac{x^2}{25}+\frac{y^2}{9}-1)\left(\frac{499}{225}\right)={(\frac{6x}{25}+\frac{4y}{9}-1)}^2$
• C. None of the above
• D. $(\frac{x^2}{25}+\frac{y^2}{9}-1)\left(\frac{499}{225}\right)=(\frac{6x}{25}+\frac{4y}{9}-1)$

Answer 1

The correct option is B $(\frac{x^2}{25}+\frac{y^2}{9}-1)\left(\frac{499}{225}\right)={(\frac{6x}{25}+\frac{4y}{9}-1)}^2$

$\text{Given equation}:\frac{x^2}{25}+\frac{y^2}{9}=1$
$\text{Here,}a=5,b=3$
We know that equation of pair of tangents from point $(x_1,y_1)$ is $S{S}_1=T^2$
$\text{Here , S}=\frac{x^2}{25}+\frac{y^2}{9}-1$
$S_1=\frac{x_1^2}{25}+\frac{y_1^2}{9}-1=\frac{36}{25}+\frac{16}{9}-1=\frac{499}{225}$

$T=\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}-1=\frac{6x}{25}+\frac{4y}{9}-1$
Substituting $S,S_1\text{and}T,$ we get:
$(\frac{x^2}{25}+\frac{y^2}{9}-1)\left(\frac{499}{225}\right)={(\frac{6x}{25}+\frac{4y}{9}-1)}^2$
Which is the required equation of the pair of tangents