JEE Advanced 2015

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Animated Solution for Mathematics - Conic Sections: If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ , then the value of $r^{2}$ is

Q: Why is the slope of the normal $-1$ at $(1, 2)$?

Let's break it down. For $y^2 = 4x$, we have $a=1$. The slope of the tangent at $(x_1, y_1)$ is $2a/y_1$. At $(1, 2)$, the tangent slope is $2(1)/2 = 1$. Since the normal is perpendicular to the tangent, its slope is the negative reciprocal of $1$, which is $-1$. It is all about that perpendicularity!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the parabola

y^{2} = 4 x

. The latus rectum is the chord passing through the focus

(a, 0)

perpendicular to the axis. For this parabola,

a = 1

, so the focus is

(1, 0)

. The endpoints of the latus rectum are

(a, 2 a)

and

(a, - 2 a)

. Can you identify these coordinates?

Hint 2

Hint 2: The Arsenal. To find the equation of the normal at a point

(x_{1}, y_{1})

on the parabola

y^{2} = 4 a x

, first find the slope of the tangent

d y / d x = 2 a / y_{1}

. The slope of the normal is the negative reciprocal of the tangent slope. Use the point-slope form

y - y_{1} = m (x - x_{1})

to find the equation of the normal line.

Hint 3

Hint 3: The Execution. The normal at

(1, 2)

x + y - 3 = 0

. The condition for this line to be a tangent to the circle

(x - 3)^{2} + (y + 2)^{2} = r^{2}

is that the perpendicular distance from the center

(3, - 2)

to the line must equal the radius

r

. Use the formula

d = ∣ A x_{1} + B y_{1} + C ∣/ A^{2} + B^{2}

to solve for

r

Visualized Solution

Visualize the Parabola $y^{2} = 4 x$

Given Parabola: $y^{2} = 4 x$
Comparing with standard form $y^{2} = 4 a x$ , we get $4 a = 4 ⟹ a = 1$ .
The focus of this parabola is at $(a, 0) = (1, 0)$ .

Identify Endpoints of Latus Rectum

The latus rectum passes through the focus $(1, 0)$ and is perpendicular to the axis.
Endpoints of Latus Rectum for $y^{2} = 4 a x$ are $(a, \pm 2 a)$ .
Substituting $a = 1$ , the endpoints are $P (1, 2)$ and $Q (1, - 2)$ .

Equation of the Normal

We need the normal at the endpoint $P (1, 2)$ .
The slope of the tangent at $(x_{1}, y_{1})$ is $\frac{d y}{d x} = \frac{2 a}{y _{1}}$ .
Therefore, the slope of the normal is $m = - \frac{y _{1}}{2 a}$ .

Slope of Normal at $P (1, 2)$

Substitute $y_{1} = 2$ and $a = 1$ into the slope formula.
$m = - \frac{2}{2 ( 1 )}$
$m = - 1$

Equation of Normal at $P (1, 2)$

Using point-slope form: $y - y_{1} = m (x - x_{1})$
$y - 2 = - 1 (x - 1)$
$y - 2 = - x + 1$
$x + y - 3 = 0$

Analyze the Circle Properties

Given Circle: $(x - 3)^{2} + (y + 2)^{2} = r^{2}$
Comparing with standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
Center $C = (3, - 2)$ and Radius is $r$ .

Apply Tangency Condition

The problem states the normal to the parabola is a tangent to this circle.
Geometric Condition: For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must exactly equal the radius $r$ .
Distance formula from $(x_{1}, y_{1})$ to $A x + B y + C = 0$ is $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ .

Setup the Distance Equation

Center $C (3, - 2)$ , Line: $x + y - 3 = 0$ .
Substitute into distance formula: $r = \frac{∣ ( 1 ) ( 3 ) + ( 1 ) ( - 2 ) - 3∣}{1 ^{2} + 1 ^{2}}$

Calculate the Radius $r$

Numerator: $∣3 - 2 - 3∣ = ∣ - 2∣ = 2$
Denominator: $1 + 1 = 2$
$r = \frac{2}{2} = 2$

Final Conclusion for $r^{2}$

We found $r = 2$ .
The question asks for the value of $r^{2}$ .
Squaring both sides: $r^{2} = (2)^{2} = 2$ .
Final Answer: $2$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We begin with the parabola defined by the equation

y^{2} = 4 x

. Comparing this to the standard form

y^{2} = 4 a x

, we identify

4 a = 4

, which yields

a = 1

Consequently, the focus of the parabola is located at

(1, 0)

The Heart of the Parabola

The latus rectum is the focal chord perpendicular to the axis of symmetry. Its endpoints are given by the coordinates

(a, 2 a)

and

(a, - 2 a)

Substituting

a = 1

, we find the critical points to be

P (1, 2)

and

Q (1, - 2)

. We proceed by constructing the normal at point

P (1, 2)

Constructing the Normal

To find the equation of the normal at

P (1, 2)

, we first determine the slope of the tangent. The derivative is given by:

\frac{d y}{d x} = \frac{2 a}{y _{1}}

At point

P (1, 2)

, the slope of the tangent is

\frac{2 ( 1 )}{2} = 1

. Since the normal is perpendicular to the tangent, its slope

m

is the negative reciprocal:

m = - 1

Using the point-slope form

y - y_{1} = m (x - x_{1})

, we substitute our values:

y - 2 = - 1 (x - 1)

Expanding this, we obtain

y - 2 = - x + 1

, which simplifies to the linear equation:

x + y - 3 = 0

The Tangency Bridge

We now consider the circle

(x - 3)^{2} + (y + 2)^{2} = r^{2}

. By comparing this to the standard form

(x - h)^{2} + (y - k)^{2} = r^{2}

, we identify the center

C

(3, - 2)

The problem states that the normal line

x + y - 3 = 0

is tangent to this circle. For a line to be tangent to a circle, the perpendicular distance from the center

C

to the line must equal the radius

r

The Final Calculation

We apply the perpendicular distance formula

d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}

r = \frac{∣ ( 1 ) ( 3 ) + ( 1 ) ( - 2 ) - 3∣}{1 ^{2} + 1 ^{2}}

Simplifying the numerator, we have

∣3 - 2 - 3∣ = ∣ - 2∣ = 2

. The denominator is

1 + 1 = 2

Thus, the radius is:

r = \frac{2}{2} = 2

Since the question asks for

r^{2}

, we calculate:

r^{2} = (2)^{2} = 2

The final result is 2.

Common Traps & FAQs

Wait, do I need to calculate the normal for both endpoints of the latus rectum?

That is a fantastic question! While you certainly can calculate both, you will find that the normals at

(1, 2)

and

(1, - 2)

are symmetric with respect to the axis of the parabola. Because the circle's center

(3, - 2)

is positioned such that both normals are equidistant from it, calculating just one is sufficient. It is a beautiful example of how symmetry can save you time in JEE problems!

Can I use the condition

c^{2} = a^{2} m^{2} + b^{2}

for the circle? I remember this from ellipses.

I love that you are connecting concepts! However, that specific condition is for the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

. For a circle, the most robust and reliable method is the geometric condition: the perpendicular distance from the center to the tangent line must equal the radius. Stick to the distance formula; it is universal and less prone to memory-based errors.

Why is the slope of the normal

- 1

(1, 2)

Let's break it down. For

y^{2} = 4 x

, we have

a = 1

. The slope of the tangent at

(x_{1}, y_{1})

2 a / y_{1}

. At

(1, 2)

, the tangent slope is

2 (1) /2 = 1

. Since the normal is perpendicular to the tangent, its slope is the negative reciprocal of

1

, which is

- 1

. It is all about that perpendicularity!

Animated Solution for Mathematics - Conic Sections: If the normals of the parabola $y^{2} = 4 x$ drawn at the end points of its latus rectum are tangents to the circle $(x - 3)^{2} + (y + 2)^{2} = r^{2}$ , then the value of $r^{2}$ is

Enter Numerical Value:

Visualized Solution

Visualize the Parabola $y^{2} = 4 x$

Identify Endpoints of Latus Rectum

Equation of the Normal

Slope of Normal at $P (1, 2)$

Equation of Normal at $P (1, 2)$

Analyze the Circle Properties

Apply Tangency Condition

Setup the Distance Equation

Calculate the Radius $r$

Final Conclusion for $r^{2}$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Heart of the Parabola

Constructing the Normal

The Tangency Bridge

The Final Calculation

Similar Questions

The point of intersection of the tangents at the ends of the latus rectum of the parabola $y^{2} = 4 x$ is.........

Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

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 two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

A triangle is formed by the tangents at the point $(2, 2)$ on the curves $y^{2} = 2 x$ and $x^{2} + y^{2} = 4 x$ , and the line $x + y + 2 = 0$ . If r is the radius of its circumcircle, then $r^{2}$ is equal to

Suppose that the normals drawn at three different points on the parabola $y^{2} = 4 x$ pass through the point $(h, k)$ . Show that $h > 2$ .

Comprehension Passage

The value of $r$ is

If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is

Enter Numerical Value:

Visualized Solution

Visualize the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=4x

Identify Endpoints of Latus Rectum

Equation of the Normal

Slope of Normal at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(1,2)

Equation of Normal at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(1,2)

Analyze the Circle Properties

Apply Tangency Condition

Setup the Distance Equation

Calculate the Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Final Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Heart of the Parabola

Constructing the Normal

The Tangency Bridge

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The point of intersection of the tangents at the ends of the latus rectum of the parabola y2=4x is.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Normals are drawn from the point P with slopes m1​,m2​,m3​ to the parabola y2=4x. If locus of P with m1​m2​=α is a part of the parabola itself then find α.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the common tangent to the parabolas, y2=4x and x2=4y also touches the circle, x2+y2=c2, then c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A tangent line L is drawn at the point (2,−4) on the parabola y2=8x. If the line L is also tangent to the circle x2+y2=a, then 'a' is equal to ,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two tangents drawn from a point P to the parabola y2=4x are at right angles, then the locus of P is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A triangle is formed by the tangents at the point (2,2) on the curves y2=2x and x2+y2=4x, and the line x+y+2=0. If r is the radius of its circumcircle, then r2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Suppose that the normals drawn at three different points on the parabola y2=4x pass through the point (h,k). Show that h>2.

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of r is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If st=1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is

Visualize the Parabola $y^{2} = 4 x$

Slope of Normal at $P (1, 2)$

Equation of Normal at $P (1, 2)$

Calculate the Radius $r$

Final Conclusion for $r^{2}$

The point of intersection of the tangents at the ends of the latus rectum of the parabola $y^{2} = 4 x$ is.........

Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

If the common tangent to the parabolas, $y^{2} = 4 x$ and $x^{2} = 4 y$ also touches the circle, $x^{2} + y^{2} = c^{2}$ , then $c$ is equal to:

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 two tangents drawn from a point $P$ to the parabola $y^{2} = 4 x$ are at right angles, then the locus of $P$ is

A triangle is formed by the tangents at the point $(2, 2)$ on the curves $y^{2} = 2 x$ and $x^{2} + y^{2} = 4 x$ , and the line $x + y + 2 = 0$ . If r is the radius of its circumcircle, then $r^{2}$ is equal to

Suppose that the normals drawn at three different points on the parabola $y^{2} = 4 x$ pass through the point $(h, k)$ . Show that $h > 2$ .

The value of $r$ is

If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is