Question

A tangent to the hyperbola $3x^2-4y^2=12$ makes equal intercepts on the axes. Then the area of the triangle formed by the tangent with coordinate axes is

• A. $2$
• B. $1/4$
• C. $4$
• D. $1/2$

Answer 1

The correct option is C $4$

Equation of tangent on $3x^2-4^2+12=0at\left(x_1,y_1\right)$ is

$3\times x{x}_1-4\times y{y}_1-12=0$

$3x{x}_1-4y{y}_1-12=0$

$\frac{3x{x}_1}{12}-\frac{4y{y}_1}{12}=1$

$\frac{x+y}{\left(4/x_1\right)\left(-3/y_1\right)}=1$

$Now$

$\left(x_1,y_1\right)$ lies on the given curve

so,

$3{x_1}^2-4{y_1}^2=12$

$3{x_1}^2-4{\left(\frac{3}{x}{x}_1\right)}^2=12-\left(1\right)$

$3{x_1}^2-4\times \frac{9}{16}{x_1}^2=12$

${x_1}^2-\frac{3}{4}{x_1}^2=4$

$4{x_1}^2-3{x_1}^2=16$

${x_1}^2=16$

$x_1=\pm 4$

$y_1=\frac{3}{4}{x}_1=\frac{3}{4}\times \pm 4=\pm 3$

$x_1=4,y_1=3or{x}_1=-4,y_1=-3$