Name: Pair of Tangent, Chord of Contact
Uploaded: 2019-07-19
Duration: 19 min 37 s
Description: This video contains all the details of the chord of the contact pair of tangent and cord bisected at a point. This concept is very useful for the students.

Question

For parabola

y^{2} = 4 a x, a > 0

. Prove that the length of the chord joining the points of contact of the tangents drawn from the point

(x_{1}, y_{1})

is

\frac{\sqrt{y_{1}^{2} + 4 a^{2}} \sqrt{y_{1}^{2} - 4 a x_{1}}}{a}

.

Accepted Answer

Equation of the parabola be given is $y^{2} = 4 a x . . . (i)$

And let $(x_{1}, y_{1})$ be any point outside the parabola.

The equation of the chord or contact from $(x_{1}, y_{1})$ to equation $(i)$ is given as,

$y y_{1} = 2 a (x + x_{1})$

$\Rightarrow x = \frac{1}{2 a} (y y_{1} - 2 a x_{1}) . . . (i i)$

To find the points of intersection, we need to solve equations $(i)$ and $(i i)$ . Hence, we get,

$y^{2} = 4 a \frac{1}{2 a} (y y_{1} - 2 a x_{1})$

$y^{2} - 2 y y_{1} + 4 a x_{1} = 0 . . . (i i i)$

Let the coordinates of the points of intersection of equations $(i)$ and $(i i)$ be $(α, β)$ and $(α^{'}, β')$ , respectively.

We can get the values of $β$ and $β'$ from equation $(i i i)$ , which can be written as,

$β + β^{'} = 2 y_{1} . . . (i v)$

And,

$β β^{'} = 4 a x_{1} . . (v)$

Again, as $(α, β)$ and $(α', β')$ are on $(i i)$ , the points will satisfy it.

So, we get,

$α = \frac{1}{2 a} (y_{1} β - 2 a x_{1})$

$α' = \frac{1}{2 a} (y_{1} β' - 2 a x_{1})$

Subtracting the above two equations, we get,

$α - α^{'} = \frac{y_{1}}{2 a} (β - β^{'}) . . . (v i)$

The distance between the points $(α, β)$ and $(α^{'}, β')$ can be given as, $\sqrt{\{{(α - α^{'})}^{2} + {(β - β^{'})}^{2}\}}$

Substituting the value of $(α - α^{'})$ from equation $(v i)$ , we get the required length of the chord as,

$= \sqrt{\frac{y_{1}^{2}}{4 a^{2}} {(β - β')}^{2} + {(β - β')}^{2}}$

$= \sqrt{({(β - β^{'})}^{2} (\frac{y_{1}^{2}}{4 a^{2}} + 1))}$

$= \frac{1}{2 a} \sqrt{(y_{1}^{2} + 4 α^{2}) ({(β + β')}^{2} - 4 β β')}$

Substituting the values from equations $(v)$ and $(i v)$ in the above equation, we get,

$= \frac{1}{2 a} \sqrt{(y_{1}^{2} + 4 α^{2}) ({(2 y_{1})}^{2} - 4 .4 a x_{1})}$

$= \frac{2}{2 a} \sqrt{(y_{1}^{2} + 4 a^{2}) (y_{1}^{2} - 4 a x_{1})}$

$= \frac{\sqrt{(y_{1}^{2} + 4 a^{2})} . \sqrt{(y_{1}^{2} - 4 a x_{1})}}{a}$

Hence proved.

S L Loney Solutions for Chapter: Conic Sections. The Parabola, Exercise 3: EXAMPLES XXVII

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: Conic Sections. The Parabola, Exercise 3: EXAMPLES XXVII

Questions from S L Loney Solutions for Chapter: Conic Sections. The Parabola, Exercise 3: EXAMPLES XXVII with Hints & Solutions

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