Question

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

• A. 3x² - 5xy - 15x + 25y = 0
• B. 3x² - 5xy + 10x + 15y = 0
• C. 2x² - 5xy - 10x + 15y = 0
• D. 2x² - 5xy - 15x + 25y = 0

Answer 1

The correct option is A

3x² - 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

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

Correct option is A