#### To determine

**To find:** The equation of the tangent line to the hyperbola at the given point.

#### Answer

The equation of the tangent line to the hyperbola at (x0,y0) is xx0a2−yy0b2=1.

#### Explanation

**Given:**

The equation of hyperbola is x2a2−y2b2=1.

The point is (x0,y0).

**Derivative rules:** *Chain rule*

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydu⋅dudx.

**Formula used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

where, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1,y=y1

**Calculation:**

Obtain the equation of the tangent line to the hyperbola at the point (x0,y0).

x2a2−y2b2=1

Differentiate the equation implicitly with respect to *x*,

ddx(x2a2−y2b2)=ddx(1)ddx(x2a2)−ddx(y2b2)=01a2ddx(x2)−1b2ddx(y2)=0

Apply the chain rule and simplify the terms,

1a2(2x)−1b2[ddy(y2)dydx]=01a2(2x)−1b2[2ydydx]=02xa2−2yb2dydx=02yb2dydx=2xa2

Multiply the equation by b22y,

dydx=2xa2×b22ydydx=b2xa2y

Thus, the derivative of the equation is dydx=b2xa2y.

That is, the slope of tangent to the equation is dydx=b2xa2y.

Therefore, the slope of tangent to the equation at (x0,y0) is m=b2x0a2y0.

Substitute (x0,y0) for (x,y) and slope m=b2x0a2y0 in equation (1),

y−y0=b2x0a2y0(x−x0)a2yy0−a2y02=b2xx0−b2x02a2yy0−b2xx=a2y020−b2x02

Divided by a2b2 on both sides,

a2yy0a2b2−b2xxa2b2=a2y02a2b2−b2x02a2b2yy0b2−xx0a2=y02b2−x02a2xx0a2−yy0b2=x02a2−y02b2

Since the point (x0,y0) on the graph of the hyperbola, x02a2−y02b2=1.

xx0a2−yy0b2=1

Therefore, the equation of the tangent line to the hyperbola at (x0,y0) is xx0a2−yy0b2=1.