Tangent to a Parabola 

Parabola is an important topic of IIT JEE Mathematics syllabus. Tangent to a parabola is an important head under parabola and it often fetches some questions in elite exams like the JEE. Hence, students are advised to prepare this topic well.

Let P(x1, y1) and Q(x2, y2) be two neighbouring points on the parabola y2 = 4ax. Then the equation of the line joining P and Q is y – y1 = (y- y1) / (x- x1) (x – x1) …… (1) 

Since, points P and Q lie on the parabola, we have

y12 = 4ax1 …… (2)

y22 = 4ax2 …… (3) 

From the above two equations we have,

y22 – y12 = 4a(x2 – x1)

⇒ (y- y1)/(x- x1) = 4a/(y+ y2)

Equation of chord PQ (i.e. equation (1) becomes):

y – y1 = 4a/(y+ y2 ) (x – x1) …… (4)

Our aim is to find the equation of tangent at point P. For that, let point Q approach point P i.e. x2 → x1 and y2 → y1.

y – y1 = 4a/(2y1) (x – x1)

yy1 = 2a (x1 + x) (using equation (2))

This is the required equation of the tangent to the parabola y2 – 4ax at P(x1, y1).

Note: 

The angle between the tangents drawn to the two parabolas at the point of their intersection is defined as the angle of intersection of two parabolas

There can be various forms of equations of tangents to a parabola. We discuss these forms one by one:

 

Tangent at the point (x1, y1) 

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

Hence, value of dy/dx at P(x1, y1) is 2a/y1 and the equation of the tangent at P is

y – y1 = 2a/y1 (x – x1) i.e. yy1 = 2a(x – x1) + y12.

⇒ yy1 = 2a(x + x1).

Alternatively, we write the equation of the chord joining the points P(x1, y1) and Q(x2, y2) on the parabola y2 = 4ax. Equation of the chord is (x – x1)/(x-x1) = (y-y1)/(y2-y1

or (x-x1)/(x2-x1) = (y-y1)(y+ y1 )/(y22-y12 )= (y-y1 )(y2+y1 )/((x2-x1 ) )

or 4a(x – x) = (y – y1) (y2 + y1).

When the two points P and Q tend to coincide, y2 → y1 and the line PQ becomes tangent to the parabola. Its equation is 4a (x – x1) = (y – y1) (2y1) = 2yy1 – 2y12= 2yy1 – 8ax1 or yy1 = 2a(x + x1). 


Tangent in terms of m 

Suppose that the equation of a tangent to the parabola y2 = 4ax … (i) 

is y = mx + c. … (ii)

The abscissae of the points of intersection of (i) and (ii) are given by the equation (mx + c)2 = 4ax. But the condition that the straight line (ii) should touch the parabola is that it should meet the parabola in coincident points 

⇒ (mx – 2a)2 = m2c2 … (iii) 

⇒ c = a/m. … (iv)

Hence, y = mx + a/m is a tangent to the parabola y2 = 4ax, whatever be the value of m.

Equation (mx + c)2 = 4ax now becomes (mx – a/m)2 = 0.

⇒ x = a/m2 and y2 = 4ax ⇒ y = 2a/m.

Thus the point of contact of the tangent y = mx + a/m is (a/m2 ,2a/m).

 

Hence the various forms of tangents are listed below:

1. yy1 = 2a(x+x1) at teh point (x1, y1)

2. y = mx + a/m

 

Illustration: 

Find the condition that the line y = mx + c may touch the Parabola y2 = 4ax and also find its point of contact. 

Solution:

Equation of Parabola is yy2 = 4ax …… (1)

Slope of tangent at any point is

dy/dx=2a/y = m (say) …… (2)

⇒ y = 2a/m

from (1), x = a/my2

⇒ point of contact is (a/my2 ,2a/m) 

Equation of tangent is

y – 2a/m = m (x-a/my2 )

or y = mx + a/m 

Therefore, the condition that y = mx + c touches the Parabola

yy2 = 4ax is c = a/m. 


Illustration: 

Find the equation of normal to the Parabola yy2 = 4ax, having slope m. 

Solution: 

Slope of tangent at any point is 

dy/dx=2a/y 

Slope of normal at that point is

 -y/2a = m (say)

⇒ Point of contact of a normal having slope ‘m’ with the Parabola

yy2 = 4ax is (amy2, – 2am) 

So, equation of normal at this point is 

y + 2am = m (x – amy2)

or y = mx – 2am – amy3


Illustration: 

If the line 2x + 3y = 1 touches the Parabola yy2 = 4ax, find the length of its latus rectum. 

Solution: 

Equation of any tangent to yy2 = 4ax is

y = mx + a/m ⇒ my2x – my + a = 0.

Comparing it with the given tangent 2x + 3y – 1 = 0, we find

my2/2=(-m)/3=a/(-1), ⇒ m = (-2)/3, a = m/3 = -2/9.

Hence the length of the latus rectum.

= 4a = 8/9, ignoring the negative sign for length. 

Alternative Solution:

Writing the given equation as

y = -2/3 x + 1/3=-2/x x-(2/9)/(-2/3), we find that a = 2/9. 

Hence the length of the latus rectum = 4a = 8/9. 

tangent at the point t 

         Equation of the tangent at ‘t’ is ty = x + aty2. T

         the point of intersection of the tangents at ‘t1’ and ‘t2’ is (at1t2, a(t1 + t2)). 

Illustration: 

         One the Parabola yy2 = 4ax, three points E, F, G are taken so that their ordinates are in G.P. Prove that the tangents at E and G intersect on the ordinate of F. 

Solution:        
        Let the points E, F, G be (at1y2, 2at1), (at2y2, 2at2), (at3y2, 2at3) respectively. Since the ordinates of these points are in G.P., t22 = t1t3tangents at E and G are t1y = x + at12 and t3y = x + at32. Eliminating y from these equation, we get x = at1t3 = at22. Hence the point lies on the ordinates of F.


Illustration: 

       Prove that the area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points.

                       parabola-matrix


The intersection of the tangents, at these points, are the points 

(at1, t2, a(t1 + t2)}, {at2t3, a(t2 + t3)}, {at3 t1, a(t3 + 1)}. 

The area of the Δ formed by these points=1/2 a2(t1 – t2) (t2 – t3) (t3 – t1). 
 

Equation of the Tangents from an External Point 

Let y = mx + a/m be any tangent to y2 = 4ax passing through the point (x1, y1).

Then, we have

y1 = mx + a/m or m2x1 – m1 + a = 0

If m1 and m2 are to roots of (i) then

m1 + m2 = y1/x1 and m1m2 = a/x1

Also the two tangents are y = m1x + a/m1 , and y = m2x + a/m2 

⇒ Their combined equation is

(y – m1x – a/m1 ) (y – m2x – a/m2 ) = 0

On solving this we get

(y2 – 4ax) (y12 – 4ax1) = [yy1 – 2a (x + x1)]2

⇒ SS1 = tWhere S = y2 – 4ax, S1 = y12 – 4x1 

Let y2 = 4ax be the equation of a parabola and (x1, y1) an external point P. Then, equation of the tangents is given by 

SS1 = t2, where S = y2 – 4ax, S1 = y12 – 4ax1, T = yy1 – 2a(x + x1). 

If the tangents from the external point (x1, y1) touch the parabola at P and Q, then PQ is the chord of contact of the tangents. 


Illustration: 

      Prove that through any given point (x1, y1) there pass, in general, two tangents to the parabola y2 = 4ax. 

Solution: 

        The equation to any tangent is y = mx + a/m. …… (1) 

       If this passes through the fixed point (x1, y1), we have 

       y1 = mx1 + a/m, i.e. m2x1 – m y1 + a = 0. …… (2) 

This is a quadratic equation and gives two values of m (real or imaginary). Corresponding to each value of m we have, two different tangents. The roots of (2) are real and different if y12 – 4ax1 > 0, i.e. If the point (x1, y1) lies outside the curve. The roots are equal, if y12 – 4ax1 = 0 i.e. if the point (x1, y1) lies on the curve. In this case the two tangent merge into one. The two roots are imaginary if y12 – 4ax1 < 0, i.e. if the point (x1, y1) lies within the curve.

 

 

Chord of Contact 


The chord joining the points of contact of the tangents on the parabola from an external point is called the chord of contact.

Let the tangent drawn from the point P(x1, y1) touch Parabola at Q(x2, y2) and R(x3, y3) then QR is the chord of contact of the point P(x1, y1) with respect to y2= 4ax.

The equation of tangents at Q and R are

yy2 = 2a(x + x2) …… (1) 

yy3 = 2a(x + x3) …… (2) 

    parabola-tangent 



Since (ii) and (iii) pass through (x1, y1) so we have

y1y2 = 2a(x1 + x2) …… (3) 

y1y3 = 2a(x1 + x3) …… (4) 

From (ii) and (iv) we find that the points Q(x2, y2) and R(x3, y3) lie on yy1 = 2a (x2+ x1), which being of first degree in x and y represents a straight line. Hence the equation of the chord of contact of P(x1, y1) is 

yy1 = 2a (x + x1) and is of the form T = 0.

Equation of the chord of contact of the tangents drawn from a point (x1, y1) to the parabola y2 = 4ax is T = 0, i.e. yy1 – 2a(x + x1) = 0. 


Note: 

The equation of the chord of the parabola y2 = 4ax with mid point 

(x1, y1) is T = S1


Illustration: 
Find the equation of the chord of the parabola y2 = 12x which is bisected at the point (5, –7).

Solution: 

Here (x1, y1) = (5, –7), and y2 = 12x = 4ax ⇒ a = 3.

The equation of the chord is S1 = T

or y12 – 4ax1 = yy1 – 2a(x + x1) or (–7)2 – 12.5 = y(–7) – 6 (x + 5). 

Or 6x + 7y + 19 = 0.

 

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