Find the equations of both the tangent lines to an ellipse that pass through a certain point?

Question

Ellipse: x^2=4*y^2=36
Point: (12,3)

Answer 1

Ellipse: x^2+4*y^2=36
Point: (12,3)

now the tangent line at a point (a,b) is given by:
2x+8yy'=0
so y'= -2x/8y = -x/4y
so for (a,b)
the slope = -a/4b
y-b=(-a/4b)(x-a)
now
this line goes through (12,3) if
3-b=(-a/4b)(12-a)
3-b= -12a/4b +a^2 / 4b
4b(3-b)= -12a+a^2
12b-4b^2= -12a+a^2
now remember that (a,b) is a point in the ellipse, so a^2+4b^2=36
so
12b+12a = a^2+4b^2 =36
so 12b+12a =36
so a+b=3, b=3-a,
now go back to the ellipse:
a^2+4b^2=36
a^2+4(3-a)^2=36
a^2+4(9-6a+a^2)=36
a^2+36-24a+4a^2=36
5a^2-24a=0,
a(5a-24)=0
which means
a=0 or a=24/5
and
b=3 or b=3-24/5
so your points are:
(0,3) and (24/5, -9/5)

Answer 2

The ellipse is
x^2 + 4y^2 = 36

The tangent at (x1, y1) is
xx1 + 4yy1=36.

This passes through (12, 3), which gives:
12 x1 + 4*3y1=36
>> x1 + y1 = 3.
>> y1 = 3 – x1

Substitute in the equation for the ellipse:
x1^2 + 4 (3 – x1)^2 = 36
>> x1^2 + 4 (x^2 - 6x1 + 9) = 36
>>5x1^2 -24x1 = 0
>> x1 = 0 or 24/5
And correspondingly, y1 = 3 or -9/5

The two tangents are

y = 3 and (24/5)x - (36/5)y = 36

Answer 3

There will only be one tangent line at a specific point.

Use implicit differentiation

2x + 8y (dy/dx) = 0

dy/dx = -2x / 8y

Slope of tangent line = (-2*12) / (8*3) = -1

y = -x + b

3 = -12 + b

b = 15

y = -x + 15