#### To determine

**To show:** The midpoint of the line segment cut from the tangent line by coordinate axes *P*.

#### Explanation

**Given:**

The equation of the hyperbola is xy=c (or)y=cx.

**Derivative rules:**

(1) Power Rule: ddx(xn)=nxn−1

(2) Constant multiple rule: ddx(cf)=cddx(f)

**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.

**Proof:**

Obtain the slope of the tangent line to the hyperbola at *P*.

The derivative of hyperbola *y* is dydx as follows,

dydx=ddx(cx)=ddx(cx−1)

Apply the constant multiple rule (2) and the power rule (1),

dydx=cddx(x−1)=c(−1x−1−1)=−cx−2=−cx2

Thus, the derivative of *y* is −cx2.

Therefore, the slope of the tangent to the hyperbola is dydx=−cx2.

Since the hyperbola is y=cx and choose the point *P* is (a,ca), the slope of the tangent line to the hyperbola at *P* is −ca2.

Substitute (a,ca) for (x1,y1) and −ca2 for *m* in equation (1),

y−ca=−ca2(x−a)y=−ca2x+ca+cay=−ca2x+2ca

Thus, the equation of tangent line at *P* is y=−ca2x+2ca.

Substitute 0 for *x* in y=−ca2x+2ca, the *y-*intercept is 2ca and the point (0,2ca).

Substitute 0 for *y* in y=−ca2x+2ca, the *x*-intercept is computed as follows.

−ca2x+2ca=0−ca2x=−2cax=2ca×a2cx=2a

Thus, the *x*-intercept of line y=−ca2x+2ca is x=2a and the point (2a,0).

The midpoint of the line segment joining the points (2a,0) and (0,2ca) is computed as follows,

(x,y)=(2a+02,0+2ca2)=(a,ca)

Therefore, it can be concluded that the midpoint of the line segment cut from the tangent line by coordinate axes is *P*.

#### To determine

**To show:** The triangle formed by the tangent line and the coordinate axis always has the same area.

#### Explanation

**Proof:**

From part (a), the *x*-intercept of the tangent line is 2*a* and the *y*-intercept of the tangent line is 2ca.

Here, the base *b* is *x*-intercept of the tangent line and the height *h* is *y*-intercept of the tangent line.

The area of the triangle bounded by axes and tangent line is computed as follows.

12bh=12(2a)(2ca)=2c

It is obvious that the area of the triangle is a constant and it is independent of *x* and *y*.

Therefore, it can be concluded that the triangle formed by the tangent line and the coordinate axes always has the same area.