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

\[{{({{x}^{2}}-{{y}^{2}})}^{3}}=4{{a}^{2}}{{x}^{2}}{{y}^{2}}\]