 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping ```Focal Chord of Parabola

Table of Content

Intersection of a Straight Line with a Parabola

Chord with a given middle point

Focal Chord of a Parabola

Length of the chord

Position of a point relative to a Parabola

Related Resources

Intersection of a Straight Line with a Parabola The combined equation of straight line y = mx + c and parabola y2 = 4ax gives us the co-ordinates of point(s) of their intersection.

The combined equation m2x2 + 2x (mc – 2a) + c2 = 0 will give those roots. The straight line therefore meets the parabola at two points.

Points of intersection of y2 = 4ax and y = mx + c are given by (mx+c)2 = 4ax i.e. m2x2 + 2x(mc – 2a) + c2 = 0. …… (i)

Since (i) is a quadratic equation, the straight line meets the parabola in two points, real, coincident, or imaginary. The roots of (i) are real or imaginary according as {2(mx – 2a)}2 – 4m2c2 is positive or negative, i.e. according as – amc + a2 is positive or negative, that is, according as mc is less than or greater than a, (taking a as positive).

Note:

When m is very small, one of the roots of equation (i) is very large; when m is equal to zero, this root is infinitely large. Hence every straight line parallel to the axis of the parabola meets the curve in one point at a finite distance and in another point at an infinite distance from the vertex. It means that a line parallel to the axis of the parabola meets the parabola only in one point.

Chord with a given middle point

Equation of the chord of the parabola y2 = 4ax whose middle point is (x1, y1) is

(y-y1) = 2a/y1(x-x1)

This can be written as T = S1, where T = yy1 – 2a(x+x1) and S1 = y12 – 4ax1.

Focal Chord of a Parabola

The chord of the parabola which passes through the focus is called the focal chord.

Any chord to y2 = 4ax which passes through the focus is called a focal chord of the parabola y2 = 4ax. Let y2 = 4ax be the equation of a parabola and (at2, 2at) a point P on it. Suppose the coordinates of the other extremity Q of the focal chord through P are (at12, 2at1).

Then, PS and SQ, where S is the focus (a, 0), have the same slopes

⇒ (2at-0)/(at2- a) = (2at1 - 0)/(at12 - a)

⇒ tt12 – t = t1t2 – t1

⇒ (tt1 + 1) (t1 – t) = 0.

Hence t1 = –1/t, i.e. the point Q is (a/t2, –2a/t).

The extremities of a focal chord of the parabola y2 = 4ax may be taken as the points t and –1/t.

Length of the chord

The abscissae of the points common to the straight line y = mx + c and the parabola y2 = 4ax are given by the equation m2x2 + (2mx – 4a) x + c2 = 0. = 4(mc-2a)2 / m4 – 4c2/m2 = 16a(a-mc)/ m4 and (y1 – y2) = m(x1-x2)

Focal Distance of a Point

The focal distance of a point P on the parabola

y2 = 4ax is the distance between the point P and the focus S, i.e. PS. Thus the focal distance of P = PS = PM = ZN = ZA + AN = a + x.

or PS = a + at2 = a(1 + t2).

Position of a point relative to a Parabola

Consider the parabola y2 = 4ax. If (x1, y1) is a given point and y12 – 4ax1 = 0, then the point lies on the parabola. But when y12 – 4ax1 ≠ 0, we draw the ordinate PM meeting the curve in L. Then P will lie outside the parabola if PM > LM, i.e., PM2 – LM2 > 0.
Now, PM2 = y12 and LM2 = 4ax1 by virtue of the coordinates of L satisfying the equation of the parabola. Hence, the condition for P to lie outside the parabola becomes y12 – 4ax1 > 0.

Similarly, the condition for P to lie inside the parabola is y12 – 4ax1 < 0.

Illustration:

Find the Length of the chord intercepted by the parabola y2 = 4ax from the line y = mx + c. Also find its mid-point.

Solution:

Simply by applying the formula of length of the line joining (x1, y1) and (x2, y2) we get,

Length of the chord = √((x1-x2)2 + (y1-y2)2 )

= √((x1-x2)2 + m2(x1-x2)2)

= |x1 – x2|√(1+m2 ) = 4√(a(a-mc)) √(1+m2 )

[ ? x1 + x2 = (-2(m-2a))/m2 and x1x2 = c2/m2 ]

The midpoint of the chord is ((2a-mc)/m2, 2a/m)

Illustration:

Prove that the circle with any focal chord of the parabola y2 = 4ax as its diameter touches its directrix.

Solution:

Let AB be a focal chord. If A is (at2, 2at), then B is (a/t2 ,-2a/t).

Equation of the circle with AB as diameter is

(x – at2) (x-a/t2) + (y – 2at) (y + 2a/t) = 0.

For x = –a, this gives (a2 (1 + t2 )2)/t2 + y2 – 2ay (t-1/t) – 4a2 = 0.

⇒ a2 (t-1/t)2 + y2 – 2ay(t – 1/t) = 0

⇒ [y – a(t – 1/t)]2 = 0, which has equal roots.

Hence x + a = 0 is a tangent to the circle with diameter AB.

Illustration:

Find the locus of the centre of the circle described on any focal chord of a parabola as diameter.

Solution:

Let the equation of the parabola be y2 = 4ax.

Let t1, t2 be the extremities of the focal chord. Then t1 . t2 = – 1.

The equation of the circle on t1, t2 as diameter is

(x – at22) (x – at22) + (y – 2at1) (y – 2at2) = 0

or x2 + y2 – ax (t12 + t22) – 2ay (t1 + t2) + a2 t12 t12 + 4a2 t1t2 = 0

⇒ x2 + y2 – ax (t12 + t22) – 2ay (t1 + t2) – 3a2 = 0. (? t1t2 = –1)

If (α,β) be the centre of the circle, then α = a/2 (t12+t22 ) If (α, β) be the centre of the circle, then α = a/2 (t12+t22)

β = a (t1 + t2) ⇒ (t1 + t2)2 = β2/a2 ⇒ t12 + t22 + 2t1t2 =β2/a2 ⇒ 2α/a-2= β2/a2

⇒ 2aα – 2a2 = β2 ⇒ β2 = 2a (α – a).

Hence locus of (α, β) is y2 = 2a(x – a).

Related Resources

For getting an idea of the type of questions asked, refer the previous year papers.

Fast Track your revision with the ultimate Revision Notes on Parabola.

To read more, Buy study materials of Parabola comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
```  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution 