Question

If tangents are drawn to the hyperbola $4x^2-5y^2=20$ parallel to the line $x-y=2$, then

• A. equation of tangent is $x-y=1$
• B. point of contact will be $(-5,-4)$
• C. point of contact will be $(5,4)$
• D. equation of tangent is $x-y+1=0$

Answer 1

Answer: (A), (B), (C), (D)

equation of tangent is $x-y+1=0$

The tangent to the hyperbola $\frac{x^2}{5}-\frac{y^2}{4}=1$ is
$y=mx\pm \sqrt{5m^2-4}$
The tangent is parallel to the line $x-y=2$ therefore, they have same slope.
$\therefore m=1$
Hence, $y=x\pm \sqrt{5-4}$
$y=x\pm 1$
$\Rightarrow y=x+1$ or $y=x-1$
$x-y+1=0$ and $x-y-1=0$
Hence point of contact will be
$(-\frac{a^{2}m}{c},-\frac{b^2}{c})\equiv (5,4),(-5,-4)$