JEE Main 2020 (9 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: If one end of focal chord $A B$ of the parabola $y^{2} = 8 x$ is at $A (1/2, - 2)$ , then the equation of tangent to it at $B$ is

Q: Wait, why is the product of parameters $t_1 t_2 = -1$ for a focal chord?

That is a fantastic question! The line passing through $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ has a slope $m = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2}{t_1 + t_2}$. If this line passes through the focus $(a, 0)$, the slope must also be $\frac{2at_1 - 0}{at_1^2 - a} = \frac{2t_1}{t_1^2 - 1}$. Equating these slopes leads directly to $t_1 t_2 = -1$. It is a beautiful geometric consequence of the parabola's definition!

Q: Can I just find the slope of the chord $AB$ and then use the point-slope form?

You absolutely can! However, using the parametric property $t_1 t_2 = -1$ is much faster and less prone to calculation errors. In competitive exams like JEE, efficiency is key. But if you prefer the slope method, go for it—just ensure your algebra is precise.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the parabola

y^{2} = 8 x

. By comparing it to the standard form

y^{2} = 4 a x

, identify the value of

a

and the coordinates of the focus

S

. Remember, a focal chord is any chord that passes through this focus.

Hint 2

Hint 2: Use the parametric form of the parabola,

(a t^{2}, 2 a t)

. Since

a = 2

, the point

A

can be written as

(2 t_{1}^{2}, 4 t_{1})

. Use the given coordinates of

A

to find the parameter

t_{1}

. Then, apply the crucial focal chord property: if the parameters of the endpoints are

t_{1}

and

t_{2}

, then

t_{1} t_{2} = - 1

Hint 3

Hint 3: Once you have determined

t_{2}

, find the coordinates of point

B

by substituting

t_{2}

back into the parametric form

(2 t_{2}^{2}, 4 t_{2})

. Finally, use the standard equation of the tangent to a parabola

y^{2} = 4 a x

at a point

(x_{1}, y_{1})

, which is

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

, to find the equation of the tangent at

B

Visualized Solution

Identify the Parabola and Focus

Given Parabola: $y^{2} = 8 x$
Standard Form: $y^{2} = 4 a x$
Comparing coefficients: $4 a = 8 ⟹ a = 2$
Focus $S = (a, 0) = (2, 0)$

Parametric Form of Point $A$

Given Point $A = (\frac{1}{2}, - 2)$
Parametric coordinates of a parabola: $(a t^{2}, 2 a t)$
Substitute $a = 2$ : $(2 t^{2}, 4 t)$

Calculate Parameter $t_{1}$ for $A$

Let parameter for $A$ be $t_{1}$
Equating y-coordinates: $4 t_{1} = - 2$
Solving for $t_{1}$ : $t_{1} = - \frac{1}{2}$

Apply Focal Chord Property

The line segment $A B$ passes through the focus $S (2, 0)$
This makes $A B$ a focal chord.
Property: If endpoints have parameters $t_{1}$ and $t_{2}$ , then $t_{1} t_{2} = - 1$

Calculate Parameter $t_{2}$ for Point $B$

Substitute $t_{1} = - \frac{1}{2}$ into the relation $t_{1} t_{2} = - 1$
$(- \frac{1}{2}) t_{2} = - 1$
Solving for $t_{2}$ : $t_{2} = 2$

Determine Coordinates of Point $B$

Point $B$ in parametric form: $(2 t_{2}^{2}, 4 t_{2})$
Substitute $t_{2} = 2$ :
$x = 2 (2)^{2} = 8$
$y = 4 (2) = 8$
Therefore, $B = (8, 8)$

Formulate the Tangent Equation

We need the tangent at point $B (8, 8)$
Equation of tangent to $y^{2} = 4 a x$ at $(x_{1}, y_{1})$ is $y y_{1} = 2 a (x + x_{1})$
Here, $a = 2$ , $x_{1} = 8$ , $y_{1} = 8$

Substitute Values into Tangent Equation

Substitute $x_{1} = 8, y_{1} = 8, a = 2$ into $y y_{1} = 2 a (x + x_{1})$
$y (8) = 2 (2) (x + 8)$
$8 y = 4 (x + 8)$

Final Equation and Conclusion

Divide the equation by 4: $2 y = x + 8$
Rearrange terms: $x - 2 y + 8 = 0$
This matches Option (3).

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given parabola is defined by the equation

y^{2} = 8 x

. By comparing this to the standard form

y^{2} = 4 a x

, we identify that

4 a = 8

, which yields

a = 2

The focus

S

of the parabola is located at

(a, 0)

, which corresponds to the point

(2, 0)

. This point serves as the anchor for our focal chord

A B

The Parametric Representation

We represent any point on the parabola using the parametric form

(a t^{2}, 2 a t)

. Substituting

a = 2

, the coordinates become

(2 t^{2}, 4 t)

For point

A (1/2, - 2)

, we equate the

y

-coordinate:

4 t_{1} = - 2 ⟹ t_{1} = - \frac{1}{2}

This parameter

t_{1} = - 1/2

uniquely defines the position of point

A

on the curve.

The Focal Chord Property

A fundamental property of a focal chord in a parabola is that the product of the parameters of its endpoints,

t_{1}

and

t_{2}

, must satisfy the relation

t_{1} t_{2} = - 1

Given

t_{1} = - 1/2

, we calculate

t_{2}

as follows:

(- \frac{1}{2}) t_{2} = - 1 ⟹ t_{2} = 2

Determining Point B

With the parameter

t_{2} = 2

identified, we find the coordinates of point

B

by substituting

t_{2}

back into the parametric form

(2 t^{2}, 4 t)

x_{B} = 2 (2)^{2} = 8

y_{B} = 4 (2) = 8

Thus, the coordinates of point

B

are

(8, 8)

The Tangent Equation

To find the equation of the tangent at

B (8, 8)

, we use the standard tangent formula for a parabola

y^{2} = 4 a x

, which is

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

. Substituting

a = 2

x_{1} = 8

, and

y_{1} = 8

y (8) = 2 (2) (x + 8)

Simplifying the expression:

8 y = 4 (x + 8)

2 y = x + 8

Rearranging the terms, we arrive at the final equation: $x - 2 y + 8 = 0$

Common Traps & FAQs

Wait, why is the product of parameters

t_{1} t_{2} = - 1

for a focal chord?

That is a fantastic question! The line passing through

(a t_{1}^{2}, 2 a t_{1})

and

(a t_{2}^{2}, 2 a t_{2})

has a slope

m = \frac{2 a t _{2} - 2 a t _{1}}{a t _{2}^{2} - a t _{1}^{2}} = \frac{2}{t _{1} + t _{2}}

. If this line passes through the focus

(a, 0)

, the slope must also be

\frac{2 a t _{1} - 0}{a t _{1}^{2} - a} = \frac{2 t _{1}}{t _{1}^{2} - 1}

. Equating these slopes leads directly to

t_{1} t_{2} = - 1

. It is a beautiful geometric consequence of the parabola's definition!

Can I just find the slope of the chord

A B

and then use the point-slope form?

You absolutely can! However, using the parametric property

t_{1} t_{2} = - 1

is much faster and less prone to calculation errors. In competitive exams like JEE, efficiency is key. But if you prefer the slope method, go for it—just ensure your algebra is precise.

Why do we use the formula

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

for the tangent?

This is the standard 'T = 0' form for the tangent to a parabola. For any curve

f (x, y) = 0

, the tangent at

(x_{1}, y_{1})

can be found by replacing

y^{2}

with

y y_{1}

and

x

with

\frac{x + x _{1}}{2}

. For

y^{2} = 4 a x

, this gives

y y_{1} = 4 a (\frac{x + x _{1}}{2}) = 2 a (x + x_{1})

. It is a universal tool that saves you from doing calculus every time!

Animated Solution for Mathematics - Conic Sections: If one end of focal chord $A B$ of the parabola $y^{2} = 8 x$ is at $A (1/2, - 2)$ , then the equation of tangent to it at $B$ is

Select Answer:

Visualized Solution

Identify the Parabola and Focus

Parametric Form of Point $A$

Calculate Parameter $t_{1}$ for $A$

Apply Focal Chord Property

Calculate Parameter $t_{2}$ for Point $B$

Determine Coordinates of Point $B$

Formulate the Tangent Equation

Substitute Values into Tangent Equation

Final Equation and Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Representation

The Focal Chord Property

Determining Point B

The Tangent Equation

Similar Questions

If one end of a focal chord $A B$ of the parabola $y^{2} = 8 x$ is at $A (\frac{1}{2}, - 2)$ , then the equation of the tangent to it at $B$ is :

The equation of a tangent to the parabola $y^{2} = 8 x$ is $y = x + 2$ . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

The equation of a tangent to the parabola, $x^{2} = 8 y$ , which makes an angle $θ$ with the positive direction of x-axis, is :

Let P be a point on the parabola, $x^{2} = 4 y$ . If the distance of P from the centre of the circle, $x^{2} + y^{2} + 6 x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at P, is :

Let $P$ be the point on the parabola, $y^{2} = 8 x$ which is at a minimum distance from the centre $C$ of the circle, $x^{2} + (y + 6)^{2} = 1$ . Then the equation of the circle, passing through $C$ and having its centre at $P$ is:

A tangent line $L$ is drawn at the point $(2, - 4)$ on the parabola $y^{2} = 8 x$ . If the line $L$ is also tangent to the circle $x^{2} + y^{2} = a$ , then 'a' is equal to ,

If $x = 9$ is the chord of contact of the hyperbola $x^{2} - y^{2} = 9$ , then the equation of the corresponding pair of tangents is

Let $C$ be the locus of the mirror image of a point on the parabola $y^{2} = 4 x$ with respect to the line $y = x$ . Then the equation of tangent to $C$ at $P (2, 1)$ is :

Tangents drawn from the point $P (1, 8)$ to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the points $A$ and $B$ . The equation of the circumcircle of the triangle $P A B$ is

If the tangents drawn to the hyperbola $4 y^{2} = x^{2} + 1$ intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one end of focal chord AB of the parabola y2=8x is at A(1/2,−2), then the equation of tangent to it at B is

Select Answer:

Visualized Solution

Identify the Parabola and Focus

Parametric Form of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Calculate Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Apply Focal Chord Property

Calculate Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2​ for Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Determine Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Formulate the Tangent Equation

Substitute Values into Tangent Equation

Final Equation and Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Representation

The Focal Chord Property

Determining Point B

The Tangent Equation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one end of a focal chord AB of the parabola y2=8x is at A(21​,−2), then the equation of the tangent to it at B is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a tangent to the parabola y2=8x is y=x+2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a tangent to the parabola, x2=8y, which makes an angle θ with the positive direction of x-axis, is :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=9 is the chord of contact of the hyperbola x2−y2=9, then the equation of the corresponding pair of tangents is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let C be the locus of the mirror image of a point on the parabola y2=4x with respect to the line y=x. Then the equation of tangent to C at P(2,1) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangents drawn from the point P(1,8) to the circle x2+y2−6x−4y−11=0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn to the hyperbola 4y2=x2+1 intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :

Animated Solution for Mathematics - Conic Sections: If one end of focal chord $A B$ of the parabola $y^{2} = 8 x$ is at $A (1/2, - 2)$ , then the equation of tangent to it at $B$ is

Parametric Form of Point $A$

Calculate Parameter $t_{1}$ for $A$

Calculate Parameter $t_{2}$ for Point $B$

Determine Coordinates of Point $B$

If one end of a focal chord $A B$ of the parabola $y^{2} = 8 x$ is at $A (\frac{1}{2}, - 2)$ , then the equation of the tangent to it at $B$ is :

The equation of a tangent to the parabola $y^{2} = 8 x$ is $y = x + 2$ . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

The equation of a tangent to the parabola, $x^{2} = 8 y$ , which makes an angle $θ$ with the positive direction of x-axis, is :

A tangent line $L$ is drawn at the point $(2, - 4)$ on the parabola $y^{2} = 8 x$ . If the line $L$ is also tangent to the circle $x^{2} + y^{2} = a$ , then 'a' is equal to ,

If $x = 9$ is the chord of contact of the hyperbola $x^{2} - y^{2} = 9$ , then the equation of the corresponding pair of tangents is

Let $C$ be the locus of the mirror image of a point on the parabola $y^{2} = 4 x$ with respect to the line $y = x$ . Then the equation of tangent to $C$ at $P (2, 1)$ is :

Tangents drawn from the point $P (1, 8)$ to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the points $A$ and $B$ . The equation of the circumcircle of the triangle $P A B$ is

If the tangents drawn to the hyperbola $4 y^{2} = x^{2} + 1$ intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is :