How to find equation of a tangent

Question

How to find equation of a tangent on a $4x^2+9y^2-24x+18y+9=0$ in $T(6,-1)$?

The solution is $x=6$, but I always get: $y = \frac{-4}{15}x + \frac {3}{5} $(?!)

Alternate form:

This is the ellipse we get: (if that helps)

Answer 1

The derivative of $4x^2+9y^2-24x+18y+9=0$ is:

$$8x+18yy'-24+18y'=0$$

or

$$4x+9yy'-12+9y'=0$$

Then,

$$ 9yy'+9y'=12-4x $$

$$9y'(y+1)=12-4x$$

Therefore, $y'=\frac{12-4x}{9(y+1)}$

Now, substitute $x=6, \ y=-1$ to find $y'$(which, as you know, is the slope a point $T$).

At this point you have everything needed finding the equation of a tangent line.

Answer 2

HINT:

We don't need to bother whether the point$(6,1)$ lies outside, on or inside the given ellipse.

Equation of any line passing through $(6,1)$ can be written as $$\frac{y-1}{x-6}=m\ \ \ \ (1)$$ where $m$ is the gradient/slope

Find the intersection of $(1)$ with the given curve by replacing $y$ with $mx-6m-1$ to form a Quadratic equation in $x$

Now for tangency the two intersection must coincide $\implies$ the discriminant of the Quadratic equation must be equal to zero

The nature of the two values of $m$ will actually determine the number of tangents

Observe:

If the point lies inside the ellipse, both the values $m$ will complex implying no real tangent

If the point lies on the ellipse, both the values $m$ will be same(real) implying exactly one real tangent

If the point lies outside the ellipse, both the values $m$ will be real & distinct implying two real real tangents