Question

Find the length of the tangent to the curve $\frac{x^2}{9}-\frac{y^2}{9}=1$ at point $(2,0)$.

• A. $\frac{5}{9}$
• B. $\frac{4}{9}$
• C. $\frac{1}{9}$
• D. None of these

Answer 1

Answer: (D)

None of these

For any standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ the equation of the tangent at point $P(x_1,y_1)$ is: $\Rightarrow \frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$

The given Hyperbola is $\frac{x^2}{9}-\frac{y^2}{9}=1,$

So equation of tangent at any point $P(x_1,y_1)$ is $\Rightarrow x{x}_1-y{y}_1=9$ ...$\left(1\right)$

As the tangent to the given hyperbola is drawn from the external point $A(2,0),$ hence the point $A(2,0)$ will lie on the tangent.

So putting the point $A(2,0)$ in equation of tangent. we get,

$\Rightarrow \left(2\right)x_1-\left(0\right)y_1=9$

$\Rightarrow x_1=\frac{9}{2}$

As point $P(x_1,y_1)$ lies also on the hyperbola, so $\frac{x_1^2}{9}-\frac{y_1^2}{9}=1,$

$\Rightarrow y_1^2=x_1^2-9$

$\Rightarrow y_1^2={\left(\frac{9}{2}\right)}^2-9$

$\Rightarrow y_1=\pm \frac{\sqrt{45}}{2}$

So point $P$ is $(\frac{9}{2},\frac{\sqrt{45}}{2})$ and $Q$ is $(\frac{9}{2},-\frac{\sqrt{45}}{2})$

As from any external point, two tangents can be drawn to hyperbola, hence one tangent meets the hyperbola at $P$ and another meets the hyperbola at $Q$. The respective length of the tangents are $AP$ and $AQ$ and $AP=AQ,$ as point $A$ lies on $x-$axis, which divides the hyperbola into two symmetrical parts.

The length of the tangent from point $A(2,0)$ is $AP$ or $AQ$, as you can see in the figure.

$AP=AQ=\sqrt{(\pm \frac{\sqrt{45}}{2}-0{)}^2+(\frac{9}{2}-2{)}^2}$

Hence length of the tangents to the given hyperbola from point $A(2,0)$ is $\frac{\sqrt{70}}{2}$