#### To determine

**To find:** The equation of the both tangent lines to the ellipse passing through the point.

#### Answer

The equation of the tangent line to the curve passing through the point (12,3) at (0,3) and (245,−95) are y=3 and y=23x−5.

#### Explanation

**Given:**

The equation of the ellipse x2+4y2=36.

The point is (12,3).

**Derivative rules:**

*Chain rule*: dydx=dydu⋅dudx.

**Calculation:**

Obtain the slope of tangent to the equation.

Differentiate x2+4y2=36 implicitly with respect to *x*.

ddx(x2+4y2)=ddx(36)ddx(x2)+4ddx(y2)=02x+4ddx(y2)=0

Apply the chain rule and simplify the terms.

2x+4[ddy(y2)dydx]=02x+4[2ydydx]=02x+8ydydx=0dydx=−x4y

Thus, the derivative of the equation of the ellipse is dydx=−x4y.

That is, the slope of the tangent is dydx=−x4y.

Let (a,b) be a point on the curve.

The slope of tangent to the curve at (a,b) is dydx=−a4b.

The equation of the tangent line passing through the point (12,3) and the slope −a4b is computed as follows.

y−3=−a4b(x−12)y−3=−ax4b+12a4b4by−12b=−ax+12a

ax+4by=12a+12b (1)

Here, the tangent line is also passing through the point (a,b).

Substitute (a,b) for (x,y),

a(a)+4b(b)=12a+12b

a2+4b2=12(a+b) (2)

The value of the curve x2+4y2=36 at (a,b) is a2+4b2=36 (3)

Substitute the equation (3) in equation (2).

12(a+b)=36a+b=3b=3−a

Substitute b=3−a in equation (3).

a2+4(3−a)2=36a2+4(9+a2−6a)=36a2+36+4a2−24a−36=05a2−24a=0

Simplify the quadratic equation,

a(5a−24)=0a=0 or a=245

Substitute a=0 in b=3−a,

Thus, the value of *b* is 3.

Substitute a=245 in b=3−a.

b=3−245=15−245=−95

Thus, the points are (0,3) and (245,−95).

Substitute (0,3) for (a,b) in equation (1),

(0)x+4y(3)=12(0)+12(3)12y=36y=3

Thus, the equation of the tangent line to the curve passing through the point (12,3) at (0,3) is y=3.

Substitute (245,−95) for (a,b) in equation (1).

(245)x+4y(−95)=12(245)+12(−95)24x−36y=288−10824x−36y=180y=23x−5

Thus, the equation of the tangent line to the curve passing through the point (12,3) at (245,−95) is y=23x−5.