Let $P$ (with $x$ coordinate $p$) be any point on the hyperbola. A tangent from $P$ strikes the $x$ axis at $(q,0)$. Prove $pq=a^2$

Question

The question is , Let $P$ (with $x$ coordinate $p$) be any point on the hyperbola $x^2/a^2-y^2/b^2 =1$. A tangent from $P$ strikes the $x$ axis at $(q,0)$. Prove $pq=a^2$

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What I have tried:

Say that the there is a point $(p,k)$ which lies on the hyperbola.

We want to find the equation of the tangent so taking the derivative of x on both sides and rearranging to find $\frac{dy}{dx}=\frac{xb^2}{ya^2}$

Hence equation of tangent is at $(y-y_1)=\frac{xb^2}{ya^2}(x-x_1)$

This tangent strikes the $x$ axis at $(q,0)$ hence the tangent equation becomes $-y_1=\frac{xb^2}{ya^2}(q-x_1)$

Now I am stuck how should I continue

Answer 1

hint...You should have $$\frac{dy}{dx}=\frac{pb^2}{ka^2}$$

So that the equation of the tangent is $$y-k=\frac{pb^2}{ka^2}(x-p)$$

Can you take it from there?

Answer 2

Be careful to keep track of which $x$’s and $y$’s are which. For the equation of the tangent line, you need the slope at $(x_1,y_1)$, so your equation for this line should start off as $$(y-y_1)={x_1b^2\over y_1a^2}(x-x_1).$$ In this equation, $x_1$ is, of course, $p$, and you can find $y_1$ by plugging this into the equation of the hyperbola. You can probably take it from there. Note that if you delay substituting for $y_1$ until you’ve manipulated this equation a bit, you might not have to solve for its value explicitly.

Here’s a slightly different, but related approach. A normal to the hyperbola at the point $(x_1,y_1)$ is given by the vector $\mathbf n=\langle x_1/a^2,-y_1/b^2\rangle$, which you can find by differentiating the equation of the hyperbola. This vector is perpendicular to the tangent at that point. We can use this normal to write a vector equation for the tangent line:$$\mathbf n\cdot\langle x,y\rangle=\mathbf n\cdot\langle x_1,y_1\rangle.$$ Plugging in the known point $(q,0)$ and setting $x_1=p$ on the left hand side results in $${pq\over a^2}={x_1^2\over a^2}-{y_1^2\over b^2}=1.$$