Point on hyperbola which is nearest to the given line Find the point on hyperbola $3x^2-4y^2=72$ which is nearest to the straight line $3x+2y+1=0$
Could someone explain me how to initiate this problem and the concept involved? 
 A: The answer is (6,-3).Just find the eqn. of tangent at a parametric point and equate its slope with that of line.Then you will easily get to the answer without much effort.You can check the image I have attached to see the solution.(https://i.stack.imgur.com/zPsCx.jp0g)
A: In $\mathbb{R}^2$, the distance from a point $(x_0,y_0)$ to a line $ax+by+c=0$ is given by the formula:
$$\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}.$$
A point on the hyperbola has coordinates $(\pm2\sqrt{y^2/3+6},y)$ with $y\in \mathbb{R}$. 
Hence you have to minimize the real functions (one for each arm):
$$f_{\pm}(y):=\frac{|\pm6\sqrt{y^2/3+6}+2y+1|}{\sqrt{10}}.$$
It turns out that the minimum distance is $f_-(3)=11/\sqrt{10}$. 
A: Every point on the Hyperbola is of the parametric form $(\frac{\sec{\theta}}{\sqrt{3}}, \frac{\tan{\theta}}{2})$
These co-ordinates can be substituted into the Line, and the Weierstraβ Substitution reduces the equation into a quadratic equation in one variable. The equation can then be explicitly solved for the exact points of intersection, which are real.
To realise that the points MUST be real, the Line given is parallel to an Asymptote of the Hyperbola. But since it is not equal to an Asymptote of the Hyperbola, the line cannot avoid the Hyperbola forever, and so, will necessarily meet it at some finite point.
A: $\frac{dy}{dx}$for the hyperbola at a point ($ x_1$,$y_1$)is $\frac{3x_1}{4y_1}$and a perpendicular to this is $\frac{-4y_1}{3x_1}$ .Perpendicular slope to the line is $\frac{2}{3}$. On equating these slopes for a line mutually perpendicular to both we get $x_1=-2y_1.$ 
Further the equation of the normal is $\frac{y-y_1}{x-x_1} = \frac{2}{3}$ and after putting $x_1=-2y_1$ we get $3y=2x +7y_1.$ 
$x=\frac{3y-7y_1}{2}.$ We can substitute this value in th equation of the hyperbola to get the point of intersection which is ($ x_1$,$y_1$). So we can put y as $y_1$ and solve for $y_1$. After this I get ($ x_1$,$y_1$) as (-6,3) and (6,-3). Substituting this point in the distance from a line formula $\frac{ax+by+c}{\sqrt{a^2+b^2}}$ I get (-6,3) as the closest point.
Is this correct?
A: I know that there are already many answers to this question, but this one does not use calculus and is really, really beautiful: 
Consider the following quadratic in $t$: 
$$\left(t\sqrt3-x\sqrt{3}\right)^2-\left(t+2y\right)^2$$
Note that this quadratic must have at least one solution, since it can be factored using the "difference of two squares" trick: 
$$\bigg[\left(1+\sqrt3\right)t-x\sqrt{3}+2y\bigg] \bigg[\left(-1+\sqrt3\right)t-x\sqrt{3}-2y\bigg]$$
It is important to consider that these solutions may not be distinct. 
Next, let's expand this quadratic: 
$$\left(t\sqrt3-x\sqrt{3}\right)^2-\left(t+2y\right)^2=2t^2-2\left(3x+2y\right)t+3x^2-4y^2$$
Since this quadratic has at least one solution, it's discriminant is greater than or equal to $0$:
$$4\left(3x+2y\right)^2-8\left(3x^2-4y^2\right)\geq 0$$ 
$$\left(3x+2y\right)^2-2\left(3x^2-4y^2\right)\geq 0$$ 
Now, let's only consider values of $x$ and $y$ that are on our hyperbola. For any of those values, $3x^2-4y^2=72$:
$$\left(3x+2y\right)^2-144\geq 0$$
$$\left(3x+2y\right)^2\geq 144$$
$$3x+2y\leq -12 \text{ or } 3x+2y\geq 12$$
Now, let's think of the geometric implications of this argument. We have deduced that the minimum of $\left|3x+2y\right|$ with constraint $3x^2-4y^2=72$ is $12$. this means that the region $\left|3x+2y\right|<12$ has no intersection with the hyperbola. It also implies that the lines $3x+2y=12$ and $3x+2y=12$ are tangent with the hyperbola. 
The original problem asks us to find the point on the hyperbola which is closest to the line $3x+2y+1=0$. The tangent at this point must be parallel to the line $3x+2y+1=0$.
So, this point is the intersection of the hyperbola with one of the two lines $3x+2y=-12$ and $3x+2y=12$. The first line intersects the hyperbola at $(-6,3)$ and the second line intersects the hyperbola at $(6,-3)$. When you compare the distances of these points from the line, you find that $(-6,3)$ is closer. You can also figure this out since $3x+2y=-12$ is closer to the line $3x+2y+1=0$ than $3x+2y=12$.

Are you wondering how I got the original quadratic in $t$? $$\left(t\sqrt3-x\sqrt{3}\right)^2-\left(t+2y\right)^2$$
I usually utilize a similar trick on ellipses with the Cauchy-Schwarz inequality. Note that the inequality I used in this problem looks much like Cauchy-Schwarz if I rearrange it:
$$\left(3x+2y\right)^2-2\left(3x^2-4y^2\right)\geq 0$$
$$\bigg[x\sqrt{3}\cdot\sqrt{3}+2y\cdot 1\bigg]^2\geq \bigg[\left(x\sqrt{3}\right)^2-(2y)^2\bigg]\bigg[\sqrt{3}^2-1^2\bigg]$$
The difference here is that there is a negative sign and the inequality sign is flipped. What I actually did in the answer above is go through a proof of Cauchy-Schwarz, but I flipped a couple of positive signs to negative signs. In the actual proof of Cauchy-Schwarz, the quadratic in $t$ is supposed to have no solutions. By flipping some signs, I made a quadratic that is guaranteed to have a solution.  
