Find equation of tangents to hyperbola

Question

$$\frac{x^2}{4} - \frac {y^2}{16} = 1$$

There is a point $(1,2)$ where $2$ lines pass through and are a tangent to both curves.

How do I find the equation of both lines?

Answer 1

Hint: The point-slope equation for the tangent line is: $y - 2 = m(x-1) \to y = mx - m + 2$, and substitute this for $y$ in the the equation of the hyperbola:

$\dfrac{x^2}{4} - \dfrac{(mx-m+2)^2}{16} = 1$. Simplify this equation as a quadratic equation in $x$, and set its delta $\triangle = 0$ to solve for $m$. You should get two values of $m$.

Answer 2

HINT:

The equation of any line passing through $(1,2)$ can be written as $$\frac{y-2}{x-1}=m\iff y=mx+2-m$$ where $m$ is the gradient

Replace the value of $y$ in the given Hyperbola Equation to form a Quadratic Equation in $x$

Now each value of $x$ represents the abscissa of the intersection(s).

For tangency the two intersections must coincide, so the roots of the Quadratic Equation in $x$ must be same.

Answer 3

HINT:

Any point on the given curve can be written as $(2\sec t,4\tan t)$

Using calculus, differentiating either sides of the given equation wrt $x,$ $$\frac{2x}4-\frac{2y}{16}\frac{dy}{dx}=0\iff\frac{dy}{dx}=\frac{4x}y$$

So, the gradient at $(2\sec t,4\tan t),$ $$ m=\dfrac{4(2\sec t)}{4\tan t}=\dfrac2{\sin t}\ \ \ \ (1)$$

Again, the gradient of the straight line passing through $(2\sec t,4\tan t);(1,2)$ is $$\frac{2-4\tan t}{1-2\sec t}\ \ \ \ (2)$$

Compare $(1),(2)$ to find $t$