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

A chord of a parabola passes through a point on the axis (outside the parabola) whose distance from the vertex is half the latus rectum. Prove that the normals at its extremities meet on the curve.

Accepted Answer

Let the equation of the parabola be $y^{2} = 4 a x$ .

The point on the axis outside the parabola at a distance of half the latus rectum from the vertex will be $(- 2 a, 0)$ .

Let $T_{1}$ and $T_{2}$ be two points on the parabola as $(a t_{1}^{2}, 2 a t_{1}) & (a t_{2}^{2}, 2 a t_{2})$ , respectively.

The line passing through $T_{1}$ and $T_{2}$ will be

$y - 2 a t_{1} = \frac{2 a t_{1} - 2 a t_{2}}{a t_{1}^{2} - a t_{2}^{2}} (x - a t_{1}^{2})$

Or $y - 2 a t_{1} = \frac{2}{t_{1} + t_{2}} (x - a t_{1}^{2})$

If this line passes through $(- 2 a, 0)$ , we get

$0 - 2 a t_{1} = \frac{2}{t_{1} + t_{2}} . (- 2 a - a t_{1}^{2})$

Or $t_{1} t_{2} = 2 \dots (i)$

Equation of the normal at $T_{1}$ is $y = - t_{1} x + 2 a t_{1} + a t_{1}^{3} \dots (i i)$

and equation of the normal at $T_{2}$ is

$y = - t_{2} x + 2 a t_{2} + a t_{2}^{3} \dots (i i i)$

To solve equations $(i i)$ and $(i i i)$ , subtracting $(i i)$ from $(i i i)$ , we get

$0 = (t_{1} - t_{2}) x - 2 a (t_{1} - t_{2}) - a (t_{1}^{3} - t_{2}^{3})$

Or $x = 2 a + a (t_{1}^{2} + t_{1} t_{2} + t_{2}^{2})$ as $t_{1} - t_{2} \neq 0$ .

Substituting this value of $x$ in $(i i)$ , we get

$y = - t_{1} [2 a + a (t_{1}^{2} + t_{1} t_{2} + t_{2}^{2})] + 2 a t_{1} + a t_{1}^{3}$

$= - a t_{1} t_{2} (t_{1} + t_{2})$

Hence, the point of intersection of the normals given by $(i i)$ and $(i i i)$ is

$[2 a + a (t_{1}^{2} + t_{1} t_{2} + t_{2}^{2}), - a t_{1} t_{2} (t_{1} + t_{2})]$

The equation of the parabola is $y^{2} - 4 a x = 0$ .

Substituting the co-ordinates of the point of intersection in the equation of the parabola,

L. H. S. $= [- a t_{1} t_{2} (t_{1} + t_{2})]^{2} - 4 a [2 a + a (t_{1}^{2} + t_{1} t_{2} + t_{2}^{2})]$

$= 4 a^{2} \{t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2}\} - 4 a^{2} \{2 + t_{1}^{2} + t_{1} t_{2} + t_{2}^{2}\} {∵ t_{1} t_{2} = 2}$

$= 4 a^{2} \{t_{1}^{2} + t_{2}^{2} + 4 - 2 - t_{1}^{2} - 2 - t_{2}^{2}\}$

$= 0 =$ R.H.S.

As the point of intersection satisfies the equation of the parabola, it lies on the curve.

Hence proved.

A chord of a parabola passes through a point on the axis (outside the parabola) whose distance from the vertex is half the latus rectum. Prove that the normals at its extremities meet on the curve.

Important Questions on Conic Sections. The Parabola

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