# Important Results on Tangents and Chords of Hyperbolas

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Consider the hyperbola \begin{align}S(x,y):\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}-1=0.\end{align}

CHORD OF CONTACT FROM P(x1, y1): From a point P(x1, y1) two tangents are drawn to the hyperbola, touching it at A and B

We wish to determine the equation of AB, the chord of contact.

We’ll follow the same procedure as we’ve been doing earlier. Suppose that A and B have the coordinates  $$({{x}_{2}},{{y}_{2}})$$ and $$({{x}_{3}},{{y}_{3}})$$ respectively. The tangents at A and B then have the equations:

\begin{align} & {{T}_{A}}:\frac{x{{x}_{2}}}{{{a}^{2}}}-\frac{y{{y}_{2}}}{{{b}^{2}}}-1=0 \\ & {{T}_{B}}:\frac{x{{x}_{3}}}{{{a}^{2}}}-\frac{y{{y}_{3}}}{{{b}^{2}}}-1=0 \\\end{align}

$${{T}_{A}}$$ and  $${{T}_{B}}$$ intersect in  $$P({{x}_{1}},{{y}_{1}}).$$ Thus,

\left. \begin{align}& \frac{{{x}_{1}}{{x}_{2}}}{{{a}^{2}}}-\frac{{{y}_{1}}{{y}_{2}}}{{{b}^{2}}}-1=0 \\ & \\& \frac{{{x}_{1}}{{x}_{2}}}{{{a}^{2}}}-\frac{{{y}_{1}}{{y}_{2}}}{{{b}^{2}}}-1=0 \\ \end{align} \right\}\,\,\,\,\,\,\,\,\,\ldots \left( 1 \right)

From the system of equations(1), we can infer that  $$({{x}_{2}},{{y}_{2}})$$ and  $$({{x}_{3}},{{y}_{3}})$$ are solutions of the linear equation

\begin{align}\boxed{\frac{{x{x_1}}}{{{a^2}}} - \frac{{y{y_1}}}{{{b^2}}} - 1 = 0}\end{align}

so that this must represent the chord of contact. The equation for the chord of contact can be written concisely as

\begin{align}\boxed{T({x_1},{y_1}) = 0}\end{align}

CHORD BISECTED P(x1, y1) : We need to find the equation of that chord of the hyperbola whose mid-point is AT $$\ P({x_1},{y_1})$$

Let that chord be QR where  $$Q\equiv ({{x}_{2}},{{y}_{2}})$$ and  $$R\equiv ({{x}_{3}},{{y}_{3}}).$$ Since Q and R lie on the hyperbola, we have

$\frac{x_{2}^{2}}{{{a}^{2}}}-\frac{y_{2}^{2}}{{{b}^{2}}}=1\\\frac{x_{3}^{2}}{{{a}^{2}}}-\frac{y_{3}^{2}}{{{b}^{2}}}=1$

Subtracting, we obtain

$\frac{({{x}_{2}}+{{x}_{3}})({{x}_{2}}-{{x}_{3}})}{{{a}^{2}}}-\frac{({{y}_{2}}+{{y}_{3}})({{y}_{2}}-{{y}_{3}})}{{{b}^{2}}}=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 1 \right)$

Since P is the mid-point of QR, we have

${{x}_{2}}+{{x}_{3}}=2{{x}_{1}},\,\,\,\,\,{{y}_{2}}+{{y}_{3}}=2{{y}_{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 2 \right)$

Using (2) in (1), we have

\begin{align}&\qquad\;\; \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{2{{x}_{1}}({{x}_{2}}-{{x}_{3}})}{{{a}^{2}}}-\frac{2{{y}_{1}}({{y}_{2}}-{{y}_{3}})}{{{b}^{2}}}=0 \\ &\qquad \Rightarrow\qquad \frac{{{y}_{2}}-{{y}_{3}}}{{{x}_{2}}-{{x}_{3}}}={{m}_{QR}}=\frac{{{b}^{2}}{{x}_{1}}}{{{a}^{2}}{{y}_{1}}} \\\end{align}

where $${{m}_{QR}}$$  denotes the slope of QR. The equation of QR can now be written using the point -slope form:

\begin{align}&\qquad\quad y-{{y}_{1}}={{m}_{QR}}(x-{{x}_{1}})\\ &\Rightarrow\quad y-{{y}_{1}}=\frac{{{b}^{2}}{{x}_{1}}}{{{a}^{2}}{{y}_{1}}}(x-{{x}_{1}}) \\ & \Rightarrow\quad \frac{x{{x}_{1}}}{{{a}^{2}}}-\frac{y{{y}_{1}}}{{{b}^{2}}}-1=\frac{x_{1}^{2}}{{{a}^{2}}}-\frac{y_{1}^{2}}{{{b}^{2}}}-1 \\ & \Rightarrow\quad \boxed{T({x_1},{y_1}) = S({x_1},{y_1})} \\\end{align}

PAIR OF TANGENTS FROM P (x1, y1) : From a point  $$P({{x}_{1}},{{y}_{1}}),$$ two tangents are drawn to the hyperbola\begin{align}\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1.\end{align}

We wish to find the joint equation of this pair of tangents. We will follow the same approach as we did in the case of the ellipse. Assume any point $$Q(h,\,k)$$ on any one of the two tangents.

The equation of PQ is therefore

\begin{align} &\qquad\; \,\frac{y-{{y}_{1}}}{x-{{x}_{1}}}=\frac{k-{{y}_{1}}}{h-{{x}_{1}}} \\\\ & \Rightarrow\quad y=\left( \frac{k-{{y}_{1}}}{h-{{x}_{1}}} \right)x+\frac{\left( h{{y}_{1}}-k{{x}_{1}} \right)}{\left( h-{{x}_{1}} \right)} \\\end{align}

If this line is a tangent to the given hyperbola, the condition for tangency

(c2 = a2m2 – b2) must be satisfied. Thus,

\begin{align}&\qquad\;\;{{\left( \frac{h{{y}_{1}}-k{{x}_{1}}}{h-{{x}_{1}}} \right)}^{2}}={{a}^{2}}{{\left( \frac{k-{{y}_{1}}}{h-{{x}_{1}}} \right)}^{2}}-{{b}^{2}} \\\\ & \Rightarrow \quad {{(h{{y}_{1}}-k{{x}_{1}})}^{2}}={{a}^{2}}{{(k-{{y}_{1}})}^{2}}-{{b}^{2}}{{(h-{{x}_{1}})}^{2}} \\\end{align}

This upon subsequent rearrangement can be written as

${{\left( \,\frac{h{{x}_{1}}}{{{a}^{2}}}-\frac{k{{y}_{1}}}{{{b}^{2}}}-1 \right)}^{2}}=\left( \frac{{{h}^{2}}}{{{a}^{2}}}-\frac{{{k}^{2}}}{{{b}^{2}}}-1 \right)\left( \frac{x_{1}^{2}}{{{a}^{2}}}-\frac{y_{1}^{2}}{{{b}^{2}}}-1 \right)$

Thus, using  $$(x,\,y)$$ instead of  $$(h,\,k),$$ we obtain the equation for the pair of tangents as

${{\left( \frac{x{{x}_{1}}}{{{a}^{2}}}-\frac{y{{y}_{1}}}{{{b}^{2}}}-1 \right)}^{2}}=\left( \,\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}-1 \right)\left( \frac{x_{1}^{2}}{{{a}^{2}}}-\frac{y_{1}^{2}}{{{b}^{2}}}-1 \right)$

which can be written concisely as

$\boxed{{T^2}({x_1},{y_1}) = S(x,y)S({x_1},{y_1})}$

Example - 21

Find the locus of the mid-points of normal chords of the hyperbola$$\;{{x}^{2}}-{{y}^{2}}={{a}^{2}}.$$

Solution: Let  $$P(h,\,k)$$ represent the mid-point of the (variable) normal chord of the given hyperbola.

It should be evident that for this normal, we can form two equations :

Equation of an arbitrary normal :  $$x\cos \theta +y\cot \theta =2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdots \left( 1 \right)$$

Chord bisected at P(h, k) : $$T(h,\,k)=0$$

$$hx-ky={{h}^{2}}-{{k}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 2 \right)$$

For (1) and (2) to represent the same line, we have

\begin{align}&\qquad\;\frac{\cos \text{ }\theta{ }}{h}=\frac{\cot \text{ }\theta{ }}{-k}=\frac{2a}{{{h}^{2}}-{{k}^{2}}} \\& \Rightarrow\quad \sec \theta=\frac{{{h}^{2}}-{{k}^{2}}}{2ah},\,\,\,\,\,\tan \text{ }\theta=-\frac{{{h}^{2}}-{{k}^{2}}}{2ak} \\\end{align}

$$\theta$$ can now be eliminated using the relation $$\text{se}{{\text{c}}^{\text{2}}}\theta -{{\tan }^{2}}\theta =1$$:

\begin{align} &\qquad\; {{\left( \frac{{{h}^{2}}-{{k}^{2}}}{2ah} \right)}^{2}}-{{\left( \frac{{{h}^{2}}-{{k}^{2}}}{2ak} \right)}^{2}}=1 \\\\ & \Rightarrow\quad {{({{h}^{2}}-{{k}^{2}})}^{3}}=4{{a}^{2}}{{h}^{2}}{{k}^{2}} \\\end{align}

Thus, the locus of the mid-point P is

${{({{x}^{2}}-{{y}^{2}})}^{3}}=4{{a}^{2}}{{x}^{2}}{{y}^{2}}$

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