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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. A line touching the parabola is said to be a tangent to the parabola provided it satisfies certain conditions. If we have a line y = mx + c touching a parabola y^{2} = 4ax, then c = a/m. Similarly, the line y = mx + c touches the parabola x^{2} = 4ay if c = -am^{2}. The line x cos c + y sin c = p touches the parabola y^{2} = 4ax if a sin^{2}c + p cos c = 0. Remark: In case, the equation of the parabola is not in standard form, then for the condition of tangency one must first try to eliminate one variable quanity out of x and y by solving the equations of straight line and the parabola and then use the condition B^{2} = 4AC for the derived quadratic equation. Equation of tangent to a parabola in Point Form Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two neighbouring points on the parabola y^{2} = 4ax. Then the equation of the line joining P and Q is y – y_{1} = (y_{2 }- y_{1}) / (x_{2 }- x_{1}) (x – x_{1}) …… (1) Since, points P and Q lie on the parabola, we have y_{1}^{2} = 4ax_{1} …… (2) y_{2}^{2} = 4ax_{2} …… (3) From the above two equations we have, y_{2}^{2} – y_{1}^{2} = 4a(x_{2} – x_{1}) ⇒ (y_{2 }- y_{1})/(x_{2 }- x_{1}) = 4a/(y_{1 }+ y_{2}) Equation of chord PQ (i.e. equation (1) becomes): y – y_{1} = 4a/(y_{1 }+ y_{2} ) (x – x_{1}) …… (4) Our aim is to find the equation of tangent at point P. For that, let point Q approach point P i.e. x_{2} → x_{1} and y_{2} → y_{1}. y – y_{1} = 4a/(2y_{1}) (x – x_{1}) ⇒ yy_{1} = 2a (x_{1} + x) (using equation (2)) This is the required equation of the tangent to the parabola y^{2} – 4ax at P(x_{1}, y_{1}). 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 in terms of m Suppose that the equation of a tangent to the parabola y^{2} = 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} = m^{2}c^{2} … (iii) ⇒ c = a/m. … (iv) Hence, y = mx + a/m is a tangent to the parabola y^{2} = 4ax, whatever be the value of m. Equation (mx + c)^{2} = 4ax now becomes (mx – a/m)^{2} = 0. ⇒ x = a/m^{2} and y^{2} = 4ax ⇒ y = 2a/m. Thus the point of contact of the tangent y = mx + a/m is (a/m^{2}, 2a/m). Hence the various forms of tangents are listed below:
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.
A line touching the parabola is said to be a tangent to the parabola provided it satisfies certain conditions. If we have a line y = mx + c touching a parabola y^{2} = 4ax, then c = a/m. Similarly, the line y = mx + c touches the parabola x^{2} = 4ay if c = -am^{2}.
The line x cos c + y sin c = p touches the parabola y^{2} = 4ax if a sin^{2}c + p cos c = 0.
Remark:
In case, the equation of the parabola is not in standard form, then for the condition of tangency one must first try to eliminate one variable quanity out of x and y by solving the equations of straight line and the parabola and then use the condition B^{2} = 4AC for the derived quadratic equation.
Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two neighbouring points on the parabola y^{2} = 4ax. Then the equation of the line joining P and Q is y – y_{1} = (y_{2 }- y_{1}) / (x_{2 }- x_{1}) (x – x_{1}) …… (1)
Since, points P and Q lie on the parabola, we have
y_{1}^{2} = 4ax_{1} …… (2)
y_{2}^{2} = 4ax_{2} …… (3)
From the above two equations we have,
y_{2}^{2} – y_{1}^{2} = 4a(x_{2} – x_{1})
⇒ (y_{2 }- y_{1})/(x_{2 }- x_{1}) = 4a/(y_{1 }+ y_{2})
Equation of chord PQ (i.e. equation (1) becomes):
y – y_{1} = 4a/(y_{1 }+ y_{2} ) (x – x_{1}) …… (4)
Our aim is to find the equation of tangent at point P. For that, let point Q approach point P i.e. x_{2} → x_{1} and y_{2} → y_{1}.
y – y_{1} = 4a/(2y_{1}) (x – x_{1})
⇒ yy_{1} = 2a (x_{1} + x) (using equation (2))
This is the required equation of the tangent to the parabola y^{2} – 4ax at P(x_{1}, y_{1}).
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:
Suppose that the equation of a tangent to the parabola y^{2} = 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} = m^{2}c^{2} … (iii)
⇒ c = a/m. … (iv)
Hence, y = mx + a/m is a tangent to the parabola y^{2} = 4ax, whatever be the value of m.
Equation (mx + c)^{2} = 4ax now becomes (mx – a/m)^{2} = 0.
⇒ x = a/m^{2} and y^{2} = 4ax ⇒ y = 2a/m.
Thus the point of contact of the tangent y = mx + a/m is (a/m^{2}, 2a/m).
Hence the various forms of tangents are listed below:
yy_{1} = 2a(x + x_{1}) at the point (x_{1}, y_{1})
y = mx + a/m at (a/m^{2}, 2a/m), provided m ≠ 0
ty = x + at^{2} at (at^{2}, 2at).
The points of intersection of the tangents at the point t_{1} and t_{2} is (at_{1}t_{2}, a(t_{1} + t_{2})).
y = mx - am^{2} is a tagnent to the parabola x^{2} = 4ay at (2am, am^{2}) for all values of m.
Some key points:
Angle between tangents at two points P(at_{1}^{2}, 2at_{1}) and Q(at_{2}^{2}, 2at_{2}) on the parabola y^{2} = 4ax is
θ = tan^{-1}|(t_{2 }- t_{1})/(1 + t_{1}t_{2})|
The orthocentre of the triangle formed by the three tangents to a parabola lies on the directrix.
It can be noted that the geometric mean of the x-coordinates of P and Q i.e. the term at_{1}t_{2} is also the x-coordinate of the point of intersection of tangents at P and Q on the parabola.
Illustration:
Find the condition that the line y = mx + c may touch the Parabola y^{2} = 4ax and also find its point of contact. Solution:
Equation of Parabola is yy^{2} = 4ax …… (1)
Slope of tangent at any point is
dy/dx=2a/y = m (say) …… (2)
⇒ y = 2a/m
from (1), x = a/my^{2}
⇒ point of contact is (a/my^{2} ,2a/m)
Equation of tangent is
y – 2a/m = m (x-a/my^{2} )
or y = mx + a/m
Therefore, the condition that y = mx + c touches the Parabola
yy^{2} = 4ax is c = a/m. Illustration:
Find the equation of normal to the Parabola yy^{2} = 4ax, having slope m.
Solution:
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
yy^{2} = 4ax is (amy^{2}, – 2am)
So, equation of normal at this point is
y + 2am = m (x – amy^{2})
or y = mx – 2am – amy^{3}.
If the line 2x + 3y = 1 touches the Parabola yy^{2} = 4ax, find the length of its latus rectum.
Equation of any tangent to yy^{2} = 4ax is
y = mx + a/m ⇒ my^{2}x – my + a = 0.
Comparing it with the given tangent 2x + 3y – 1 = 0, we find
my^{2}/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.
One the parabola yy^{2} = 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.
Let the points E, F, G be (at_{1}y^{2}, 2at_{1}), (at_{2}y^{2}, 2at_{2}), (at_{3}y^{2}, 2at_{3}) respectively. Since the ordinates of these points are in G.P., t_{2}^{2} = t_{1}t_{3}. tangents at E and G are t_{1}y = x + at_{1}2 and t_{3}y = x + at_{3}^{2}. Eliminating y from these equation, we get x = at_{1}t_{3} = at_{2}^{2}. Hence the point lies on the ordinates of F.
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. The intersection of the tangents, at these points, are the points (at_{1}, t_{2}, a(t_{1} + t_{2})}, {at_{2}t_{3}, a(t_{2} + t_{3})}, {at_{3} t_{1}, a(t_{3} + 1)}. The area of the Δ formed by these points=1/2 a2(t_{1} – t_{2}) (t_{2} – t_{3}) (t_{3} – t_{1}).
Let y = mx + a/m be any tangent to y^{2} = 4ax passing through the point (x_{1}, y_{1}).
Then, we have
y_{1} = mx + a/m or m_{2}x_{1} – m_{1} + a = 0
If m_{1} and m_{2} are to roots of (i) then
m_{1} + m_{2} = y_{1}/x_{1} and m_{1}m_{2} = a/x_{1}
Also the two tangents are y = m_{1}x + a/m_{1} , and y = m_{2}x + a/m_{2}
⇒ Their combined equation is
(y – m_{1}x – a/m_{1} ) (y – m_{2}x – a/m_{2} ) = 0
On solving this we get
(y^{2} – 4ax) (y_{1}^{2} – 4ax_{1}) = [yy_{1} – 2a (x + x_{1})]^{2}
⇒ SS_{1} = T^{2}_{ }Where S = y^{2} – 4ax, S_{1} = y_{1}^{2} – 4x_{1}
Let y^{2} = 4ax be the equation of a parabola and (x_{1}, y_{1}) an external point P. Then, equation of the tangents is given by
SS_{1} = T^{2}, where S = y^{2} – 4ax, S_{1} = y_{1}^{2} – 4ax_{1}, T = yy_{1} – 2a(x + x_{1}).
If the tangents from the external point (x_{1}, y_{1}) touch the parabola at P and Q, then PQ is the chord of contact of the tangents. Illustration:
Prove that through any given point (x_{1}, y_{1}) there pass, in general, two tangents to the parabola y^{2} = 4ax.
The equation to any tangent is y = mx + a/m. …… (1)
If this passes through the fixed point (x_{1}, y_{1}), we have
y_{1} = mx_{1} + a/m, i.e. m_{2}x_{1} – m y_{1} + 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 y_{1}^{2} – 4ax_{1} > 0, i.e. If the point (x_{1}, y_{1}) lies outside the curve. The roots are equal, if y_{1}^{2} – 4ax_{1} = 0 i.e. if the point (x_{1}, y_{1}) lies on the curve. In this case the two tangent merge into one. The two roots are imaginary if y_{1}^{2} – 4ax_{1} < 0, i.e. if the point (x_{1}, y_{1}) lies within the curve.
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(x_{1}, y_{1}) touch Parabola at Q(x_{2}, y_{2}) and R(x_{3}, y_{3}) then QR is the chord of contact of the point P(x_{1}, y_{1}) with respect to y^{2}= 4ax.
The equation of tangents at Q and R are
yy_{2} = 2a(x + x_{2}) …… (1)
yy_{3} = 2a(x + x_{3}) …… (2) Since (ii) and (iii) pass through (x_{1}, y_{1}) so we have
y_{1}y_{2} = 2a(x_{1} + x_{2}) …… (3)
y_{1}y_{3} = 2a(x_{1} + x_{3}) …… (4) From (ii) and (iv) we find that the points Q(x_{2}, y_{2}) and R(x_{3}, y_{3}) lie on yy_{1} = 2a (x_{2}+ x_{1}), which being of first degree in x and y represents a straight line. Hence the equation of the chord of contact of P(x_{1}, y_{1}) is
yy_{1} = 2a (x + x_{1}) and is of the form T = 0.
Equation of the chord of contact of the tangents drawn from a point (x_{1}, y_{1}) to the parabola y^{2} = 4ax is T = 0, i.e. yy_{1} – 2a(x + x_{1}) = 0.
The locus of point of intersection of tagent to the parabola y^{2} = 4ax with angle between them as θ is given by y^{2} – 4ax = (a + x)^{2} tan^{2}θ.
If this angle θ = 0° or 180°, then the locus is y^{2 }- 4ax = 0 which is actually the given parabola itself.
but, if θ = 90°, then the locus is x + a = 0 which is the equation of the directrix of the parabola. Note:
The equation of the chord of the parabola y^{2} = 4ax with mid point (x_{1}, y_{1}) is T = S_{1}. Illustration:
Find the equation of the chord of the parabola y2 = 12x which is bisected at the point (5, –7).
Here (x_{1}, y_{1}) = (5, –7), and y^{2} = 12x = 4ax ⇒ a = 3.
The equation of the chord is S_{1} = T
or y_{1}^{2} – 4ax_{1} = yy_{1} – 2a(x + x_{1}) or (–7)2 – 12.5 = y(–7) – 6 (x + 5).
Or 6x + 7y + 19 = 0.
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