Question

The equation of the pair of tangents drawn from the point (4,3) to the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ is .

• A. $3x^2-4\mathrm{xy}-12x+16y=0$
• B. $3x^2-2\mathrm{xy}-12x+16y=0$
• C. $3x^2-8\mathrm{xy}-12x+16y=0$
• D. $3x^2+4\mathrm{xy}-12x+16y=0$

Answer 1

The correct option is A $3x^2-4\mathrm{xy}-12x+16y=0$

The equation of the pair of tangents is given by:
${\mathrm{SS}}_1=T^2\Rightarrow (\frac{x^2}{16}-\frac{y^2}{9}-1)(-1)=(\frac{4x}{16}-\frac{3y}{9}-1{)}^2\Rightarrow -\frac{x^2}{16}+\frac{y^2}{9}+1=\frac{x^2}{16}+\frac{y^2}{9}+1-\frac{\mathrm{xy}}{6}-\frac{x}{2}+\frac{2y}{3}\Rightarrow \frac{x^2}{8}-\frac{\mathrm{xy}}{6}-\frac{x}{2}+\frac{2y}{3}=0\Rightarrow 3x^2-4\mathrm{xy}-12x+16y=0$