#### To determine

**To Find:** the values xT,yT, xN,yN can take.

#### Answer

xT takes values from (3, ∞) and yT takes values from (2, ∞).

xN takes values from ( 0,53 ) and yN takes values from (-52, 0).

#### Explanation

**1) Concept:**

We use implicit differentiation to find the equation of tangent and normal line .

**2) Given:**

The ellipse

x29+y24=1T and N are tangent and normal lines at point P in the first quadrant. The figure is as shown below.

xT and yT are x and y intercepts of T and xN and yN are x and y intercepts of N.

We can say that as point P approaches to (0, 2), xT approachesto∞and yT approaches to 2. As P approaches to (0, 3) from right, xT approaches* to* 3* and *yT approaches to ∞.

So we may guess that that xT is in interval (3, ∞) and yT is in (2, ∞)xN is in interval (0, 3) and yN is in (-∞, 0).

**3) Calculations:**

x29+y24=1

Taking implicit differentiation,

2x9+2y4y'=0

So y'=-49xy

Let (x0,y0) be a point on the ellipse.

The equation of tangent line is

y-y0= -4x09y0 ( x-x0)

It simplifies to

4x0x+9 y0y=4 x02+9 yx02

Divide it by 36.

x0x9+y0y4=x029+y024=1

Thus the equation of tangent line is

x0x9+y0y4=1

The x intercept xT for the tangent line is given by

xTx9=1

So

xT=9x0

and y intercept

yT=4y0

So as x_{0} takes value in (0, 3) xT takes values from (3, ∞), and as y0 takes value in (0, 2)yT takes values from (2, ∞).

Now we find equation of normal line.

At point x0 ,y0 the slope of normal line is

-1y'x0,y0=9y04x0

The equation is

y-y0=9y04x0 ( x-x0)

So x intercept xN is given by:

0-y0=9y04x0 ( xN-x0)

XN= -4x09+x0=5x09

Y intercept yN is given by yn-y0=9y04x0 ( 0-x0)

YN= -9y04+y0= -5y04

So as x0 takes value in (0 ,3) xN takes values from (0, 5/3), and as y_{0} takes value in (0,2),yN takes values from (-5/2, 0). The guess was very off about yN.

**Conclusion:**

xT takes values from (3, ∞) and yT takes values from (2, ∞).

xN takes values from ( 0,53 ) and yN takes values from (-52, 0).