Question

Find the equation of the tangents drawn form the point (5,3) to the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1.$

• A. $3x^2-5xy-15x+25y=0$
• B. $3x^2-5xy+10x+15y=0$
• C. $2x^2-5xy-10x+15y=0$
• D. $2x^2-5xy-15x+25y=0$

Answer 1

The correct option is A

$3x^2-5xy-15x+25y=0$

Given point is p(5,3)

Hyperbola $S=\frac{x^2}{25}-\frac{y^2}{9}-1=0$

$\because S_1=\frac{x^2}{25}-\frac{y^2}{9}-1=-1<0$

$\Rightarrow$ Point $P\equiv (5,3)$ lies outside the hyperbola

$\therefore$ Two tangents can be drawn from the point p(5,3) and

equation of pair of pair of tangents is $S{S}_1=T^2$

$(\frac{x^2}{25}-\frac{y^2}{9}-1)(-1)={(\frac{5x}{25}-\frac{2y}{9}-1)}^2$

$\frac{-x^2}{25}+\frac{y^2}{9}+1=\frac{x^2}{25}+\frac{y^2}{9}+1-\frac{-2xy}{15}+\frac{2y}{3}-\frac{2x}{5}$

$\frac{2x^2}{25}-\frac{2xy}{15}+\frac{2y}{3}-\frac{2x}{5}=0$

$3x^2-5xy+25y-15x=0$

Equation of pair of tangents

$3x^2-5xy-15x+25y=0$