Question

Through the positive vertex of a hyperbola a tangent is drawn; where does it meet the conjugate hyperbola?

Answer 1

Hint: We here will first assume the hyperbola to be $H_1$ given as $${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$$. Then, we will calculate its conjugate by either reversing the signs of the coefficients of $x^2$ and $y^2$or by multiplying LHS or RHS by ‘-1’ and name it $H_2$. Then we will draw the figure of this hyperbola along with its tangent and calculate its positive vertex. Then we will calculate the tangent at that point given as:
The tangent at any point $(h,k)$ on a curve f(x)=y is given as:
$y-k={\left({\displaystyle \frac{dy}{dx}}\right)}_{(h,k)}(x-h)$
Then we will solve the each calculated equation of the conjugate hyperbola $H_2$ and hence we will obtain the required points of intersection.

Complete step-by-step solution:
Here we have been given a hyperbola. Let us first assume the hyperbola to be $H_1$ given as $${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$$. Now, we know that in a conjugate hyperbola, the signs of the coefficients of $x^2$ and $y^2$ are reversed, or we can say that either the LHS or the RHS is multiplied by ‘-1’ to obtain the equation of the conjugate hyperbola. Let us assume the conjugate hyperbola to be $H_2$. Hence, $H_2$ is given as:
$H_2\equiv {\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=-1$
Now, we will first consider the hyperbola $H_1\equiv {\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$
We know that in this type of hyperbola, there are two vertices given as $V_1(a,0)$ and $V_2(-a,0)$. We also know that the respective foci of this hyperbola is given as $F_1(ae,0)$ and $F_2(-ae,0)$
This hyperbola is shown as follows:

Now, we can see that the positive vertex of this hyperbola is $V_1(a,0)$. Now, the tangent on this point is given as follows:
We know that the tangent at any point $(h,k)$ on a curve f(x)=y is given as:
$y-k={\left({\displaystyle \frac{dy}{dx}}\right)}_{(h,k)}(x-h)$
Here, we have the hyperbola as ${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$
Now, the value of ${\displaystyle \frac{dy}{dx}}$ is given as:
${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$
Differentiating w.r.t. x on both sides we get:
$\begin{array}{rl}& {\displaystyle \frac{2x}{a^2}}-{\displaystyle \frac{2y}{b^2}}{\displaystyle \frac{dy}{dx}}=0\\ & \Rightarrow {\displaystyle \frac{2x}{a^2}}={\displaystyle \frac{2y}{b^2}}{\displaystyle \frac{dy}{dx}}\\ & \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{x}{y}}{\displaystyle \frac{b^2}{a^2}}\end{array}$
 Now, the value of ${\displaystyle \frac{dy}{dx}}$ at (a,0) is given as:
$\begin{array}{rl}& {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{x}{y}}{\displaystyle \frac{b^2}{a^2}}\\ & \Rightarrow {\left({\displaystyle \frac{dy}{dx}}\right)}_{(a,0)}={\displaystyle \frac{a}{0}}{\displaystyle \frac{b^2}{a^2}}\\ & \Rightarrow {\left({\displaystyle \frac{dy}{dx}}\right)}_{(a,0)}={\displaystyle \frac{1}{0}}\end{array}$
Thus, the required tangent is given as:
$\begin{array}{rl}& y-0={\displaystyle \frac{1}{0}}(x-a)\\ & \Rightarrow 0\left(y\right)=1(x-a)\\ & \Rightarrow x-a=0\\ & \Rightarrow x=a\end{array}$
 Now, as mentioned above, the conjugate hyperbola is $H_2\equiv {\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=-1$
We will now solve the tangent with this hyperbola to obtain the required points of intersection.
We here have the tangent as
$x=a$ …..(i)
Now, putting this value of x in $H_2$ and solving we get:
$\begin{array}{rl}& {\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=-1\\ & \Rightarrow {\displaystyle \frac{{\left(a\right)}^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=-1\\ & \Rightarrow {\displaystyle \frac{a^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=-1\\ & \Rightarrow 1-{\displaystyle \frac{y^2}{b^2}}=-1\\ & \Rightarrow 2={\displaystyle \frac{y^2}{b^2}}\\ & \Rightarrow y^2=2b^2\end{array}$
Now, taking under root on both sides, we get:
$\begin{array}{rl}& y^2=2b^2\\ & \Rightarrow y=\pm \sqrt{2}b\end{array}$
Thus, the required points of intersections are $(a,\pm \sqrt{2}b)$.

Note: We can also obtain the equation of the tangent at a point (h,k) on any curve of degree 2 by using the method of keeping T=0. In this method, the variables with power 2 change as:
$\begin{array}{rl}& x^2\to h.x\\ & y^2\to k.y\end{array}$
The variables of power 1 change as:
$\begin{array}{rl}& x\to {\displaystyle \frac{1}{2}}(x+h)\\ & y\to {\displaystyle \frac{1}{2}}(y+k)\end{array}$
The constants in this method remain the same.
If we find the equation of tangent on $H_1$ by using T=0 at (a,0) we get:
$\begin{array}{rl}& {\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1\\ & \Rightarrow {\displaystyle \frac{a.x}{a^2}}-{\displaystyle \frac{0.y}{b^2}}=1\\ & \Rightarrow {\displaystyle \frac{x}{a}}-0=1\\ & \Rightarrow x=a\end{array}$