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Important Results on Tangents and Chords of Parabolas

Go back to  'Parabola'

In the unit on circles, we derived some significant results:

Pair of tangents from \(({x_1},{y_1})\)

:

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

Chord of contact of the two tangents drawn from\(({x_1},{y_1})\)

:

\(T({x_1},{y_1}) = 0\)

Chord bisected at a given point \(({x_1},{y_1})\)

:

\(T({x_1},{y_1}) = S({x_1},{y_1})\)

In this section we’ll see that these results can be generalised to the parabola in the same forms !

Result - 1 : We first of all prove that two real and distinct tangents can be drawn from any external point to a given parabola. Let the parabola \({y^2} = 4ax\) and P(h, k) be an external point so that \({k^2} > 4ah.\)

The equation of any tangent to the parabola can be written as

\[y = mx + \frac{a}{m}\]

If this passes through (h, k), we have

\[\begin{align}& \qquad k = mh + \frac{a}{m}\\\\&\Rightarrow \quad h{m^2} - km + a = 0\end{align}\]

This quadratic obviously has two real and distinct values of m as roots since the discriminant, \({k^2} - 4ah,\) is positive.

This analysis also confirms the trivial result that from any point inside a parabola, no tangent can be drawn onto it.

Result - 2 Pair of tangents from P (h, k) : We wish to determine the joint equation of the pair of tangents drawn from an external point P (h, k) to the parabola \({y^2} = 4ax.\)

 The equation of any tangent to \({y^2} = 4ax\) can be written as \(\begin{align}y = mx + \frac{a}{m}\end{align}\)

If this passes through P(h, k), we must have

\[\begin{align}& \qquad k = mh + \frac{a}{m}\\\\&\Rightarrow \quad {m^2}h - km + a = 0\end{align}\]

Let \({m_1}\;\;\text{and}\;\;{m_2}\) be the two roots of this quadratic. Thus,

\[{m_1} + {m_2} = \frac{k}{h},\,\,{m_1}{m_2} = \frac{a}{h} \qquad \qquad\qquad\dots\left( 1 \right)\]

The joint equation can now be written as

\[\begin{align}& \qquad \left( {y - {m_1}x - \frac{a}{{{m_1}}}} \right)\left( {y - {m_2}x - \frac{a}{{{m_2}}}} \right) = 0\\\\&{y^2} + {m_1}{m_2}{x^2} - ({m_1} + {m_2})xy + a\left( {\frac{{{m_1}}}{{{m_2}}} + \frac{{{m_2}}}{{{m_1}}}} \right)x - \left( {\frac{a}{{{m_1}}} + \frac{a}{{{m_2}}}} \right)y + \frac{{{a^2}}}{{{m_1}{m_2}}} = 0\end{align}\]

All the coefficients can be expressed in terms of \(({m_1} + {m_2}) \,and \,({m_1}{m_2}).\) Thus, using (1), we obtain

\[\begin{align}& \qquad {y^2} + \frac{{a{x^2}}}{h} - \frac{{kxy}}{h} + \left( {\frac{{{k^2} - 2ah}}{h}} \right)x - ky + \frac{h}{a} = 0\\\\&\Rightarrow a{(x - h)^2} = (y - k)(xk - yh)\end{align}\]

This can be written as

\[({y^2} - 4ax)({k^2} - 4ah) = {\{ yk - 2a(x + h)\} ^2} \qquad\qquad\qquad \dots \left( 2 \right)\]

Now, if we denote \({{y}^{2}}-4ax\;\;\text{as }\;\;S(x,y)\text{ }and\text{ }yk-2a(x+h)\;\;\text{as }\;\text{ }T(h,k),\) this relation we obtained in (2) can be written concisely as

\[S(x,y)S(h,k) = {(T(h,k))^2}\]

which is the same form as that of the circle. Thus, we can write the pair of tangents from (h, k) as

\[\fbox{\(\begin{array}{*{20}{c}} {{{(T(h,k))}^2} = S(x,y)S(h,k)} \end{array}\)}\]

which is sometimes more simply written as \({T^2} = S{S_1}.\)

Chord of contact from P(h, k): To evaluate the chord of contact from P(h, k) to the parabola \({y^2} = 4ax,\) we’ll follow the same approach as we did for circles. Let the tangents from P to the parabola touch it at \(A({x_1},{y_1})\;\; and\;\; B({x_2},{y_2}).\) We can, using the equation of tangent at a point, write the equations of both PA and PB as

\[\begin{align}&y{y_1} = 2a(x + {x_1})\\\\&y{y_2} = 2a(x + {x_2})\end{align}\]

Since both these lines pass through P(h, k), we have

\[\begin{align}&k{y_1} = 2a(h + {x_1}) \qquad \dots\left( 1 \right)\\\\&k{y_2} = 2a(h + {x_2}) \qquad \dots \left( 2 \right)\end{align}\]

Now comes the crucial step; from (1) and (2), we can conclude that both \(A({x_1},{y_1})\;and\;B({x_2},{y_2})\) satisfy the linear equation

\[ky = 2a(h + x)\]

which must then be what we are looking for: the chord of contact.

As in the case of circles, the equation to the chord of contact can be written concisely as

\[\fbox{\(\begin{array}{*{20}{c}} {T(h,k) = 0} \end{array}\)}\]

Chord bisected at (h, k) : Any line passing through (h, k) can be written as

\[y - k = m(x - h) \qquad \dots \left( 1 \right)\]

The intersection of this line with the parabola will, in general, give two points, say \(A({x_1},{y_1})\;and\;B({x_2},{y_2})\) The co-ordinates of A and B can be obtained by simultaneously solving the equation of this line with that of the parabola, i.e. \({y^2} = 4ax.\)

Thus

\[{(m(x - h) + k)^2} = 4ax\]

\[ \Rightarrow {m^2}x + (2m(k - mh) - 4a)x + {(k - mh)^2} = 0\]

The roots of this equation will be \({x_1}\;and\;{\rm{ }}{x_2}\) so that

\[{x_1} + {x_2} = \frac{{4a - 2m(k - mh)}}{{{m^2}}}\]

This must equal 2h, since h is the mid-point of A and B. Thus,

\[\begin{align}&\qquad \;2h = \frac{{4a - 2mk + 2{m^2}h}}{{{m^2}}}\\&\Rightarrow \quad m = \frac{{2a}}{k}\end{align}\]

Substituting this in (1), we get the required equation of the chord as

\[\begin{align}& \qquad \;\;y - k = \frac{{2a}}{k}(x - h)\\\\&\Rightarrow \quad yk - {k^2} = 2a(x - h)\end{align}\]

We write this as

\[yk - 2a(x + h) = {k^2} - 4ah\] because now this equation has the concise representation

\[\fbox{\(\begin{array}{*{20}{c}} {T(h,k) = S(h,k)} \end{array}\)},\]the same as in the case of circles.

Download SOLVED Practice Questions of Important Results on Tangents and Chords of Parabolas for FREE
Parabolas
grade 11 | Questions Set 1
Parabolas
grade 11 | Answers Set 1
Parabolas
grade 11 | Questions Set 2
Parabolas
grade 11 | Answers Set 2
Download SOLVED Practice Questions of Important Results on Tangents and Chords of Parabolas for FREE
Parabolas
grade 11 | Questions Set 1
Parabolas
grade 11 | Answers Set 1
Parabolas
grade 11 | Questions Set 2
Parabolas
grade 11 | Answers Set 2
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