JEE Main 2016

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: 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:

Q: Why do we use the normal? Can't we just use the distance formula and minimize it with calculus?

You absolutely can! Using the distance formula $D^2 = (x-0)^2 + (y+6)^2$ and substituting $y^2=8x$ would lead to a function of $x$ that you could differentiate. However, that approach often leads to messy algebra. The normal method is the 'geometric shortcut'—it's elegant, faster, and less prone to calculation errors. In JEE, time is your currency; spend it wisely!

Select Answer:

Visualized Solution

Visualizing the Problem

Given parabola: $y^{2} = 8 x$ with $a = 2$
Given circle: $x^{2} + (y + 6)^{2} = 1$
Center of the given circle: $C (0, - 6)$
We need to find a point $P$ on the parabola at a minimum distance from $C$ .

The Minimum Distance Principle

Key Geometric Concept: The shortest distance from an external point to a smooth curve always lies along the normal to the curve.
Therefore, the normal to the parabola at point $P$ must pass through the center $C (0, - 6)$ .

Parametric Representation of $P$

For the parabola $y^{2} = 4 a x$ , any parametric point is $P (a t^{2}, 2 a t)$ .
Comparing $y^{2} = 8 x$ with $y^{2} = 4 a x$ , we get $4 a = 8 \Rightarrow a = 2$ .
Thus, the parametric coordinates of $P$ are $P (2 t^{2}, 4 t)$ .

Equation of the Normal

The equation of the normal to the parabola $y^{2} = 4 a x$ in terms of parameter $t$ is:
$y = - t x + 2 a t + a t^{3}$
Substituting $a = 2$ , we get:
$y = - t x + 4 t + 2 t^{3}$

Normal Passing Through $C (0, - 6)$

Since the normal passes through the center $C (0, - 6)$ , substitute these coordinates into the normal equation:
$- 6 = - t (0) + 4 t + 2 t^{3}$
This simplifies to:
$- 6 = 4 t + 2 t^{3}$

Solving the Cubic Equation

Rearranging the terms:
$2 t^{3} + 4 t + 6 = 0$
Divide by $2$ to simplify:
$t^{3} + 2 t + 3 = 0$
By inspection, $t = - 1$ is a root because $(- 1)^{3} + 2 (- 1) + 3 = - 1 - 2 + 3 = 0$ .

Finding Coordinates of $P$

Substitute $t = - 1$ back into the parametric coordinates of $P (2 t^{2}, 4 t)$ :
$x = 2 (- 1)^{2} = 2$
$y = 4 (- 1) = - 4$
Thus, the point of minimum distance is $P (2, - 4)$ .

Calculating the Radius $R$

The new circle is centered at $P (2, - 4)$ and passes through $C (0, - 6)$ .
The radius $R$ is the distance $P C$ :
$R^{2} = (2 - 0)^{2} + (- 4 - (- 6))^{2}$
$R^{2} = 2^{2} + 2^{2} = 4 + 4 = 8$

Equation of the New Circle

Using the standard circle equation $(x - h)^{2} + (y - k)^{2} = R^{2}$ with center $(2, - 4)$ and $R^{2} = 8$ :
$(x - 2)^{2} + (y + 4)^{2} = 8$
Expanding the terms:
$x^{2} - 4 x + 4 + y^{2} + 8 y + 16 = 8$
$x^{2} + y^{2} - 4 x + 8 y + 12 = 0$
This matches Option 3.

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are given a parabola defined by the equation

y^{2} = 8 x

and a circle defined by

x^{2} + (y + 6)^{2} = 1

. Our objective is to identify the point

P

on the parabola that is closest to the center of the circle,

C (0, - 6)

The geometric principle at play is that the shortest distance from a point to a curve occurs along the normal to the curve at that point. Therefore, the line segment

C P

must be perpendicular to the tangent of the parabola at

P

Parametric Representation

For a parabola of the form

y^{2} = 4 a x

, we identify

4 a = 8

, which implies

a = 2

. We can represent any point

P

on this parabola using the parametric coordinates:

P = (a t^{2}, 2 a t) = (2 t^{2}, 4 t)

The general equation for the normal to the parabola at point

t

is given by:

y = - t x + 2 a t + a t^{3}

Substituting

a = 2

into this equation, we obtain the specific normal equation:

y = - t x + 4 t + 2 t^{3}

Solving for the Parameter

Since the normal must pass through the center of the circle

C (0, - 6)

, we substitute

x = 0

and

y = - 6

into the normal equation:

- 6 = - t (0) + 4 t + 2 t^{3}

2 t^{3} + 4 t + 6 = 0

Dividing the entire equation by

2

, we simplify the expression to:

t^{3} + 2 t + 3 = 0

By inspection, we test for roots and find that

t = - 1

satisfies the equation:

(- 1)^{3} + 2 (- 1) + 3 = - 1 - 2 + 3 = 0

Final Calculation

Using the value

t = - 1

, we determine the coordinates of point

P

P = (2 (- 1)^{2}, 4 (- 1)) = (2, - 4)

The point on the parabola closest to the center of the circle is $(2, - 4)$ .

To verify the distance, we calculate the radius squared

R^{2}

from

P (2, - 4)

C (0, - 6)

R^{2} = (2 - 0)^{2} + (- 4 - (- 6))^{2} = 2^{2} + 2^{2} = 8

The resulting circle centered at

P

with this radius is:

(x - 2)^{2} + (y + 4)^{2} = 8

x^{2} + y^{2} - 4 x + 8 y + 12 = 0

Common Traps & FAQs

Why do we use the normal? Can't we just use the distance formula and minimize it with calculus?

You absolutely can! Using the distance formula

D^{2} = (x - 0)^{2} + (y + 6)^{2}

and substituting

y^{2} = 8 x

would lead to a function of

x

that you could differentiate. However, that approach often leads to messy algebra. The normal method is the 'geometric shortcut'—it's elegant, faster, and less prone to calculation errors. In JEE, time is your currency; spend it wisely!

How do I know which root of the cubic equation to pick?

Great question! A cubic equation can have up to three real roots. In this specific problem, we found

t^{3} + 2 t + 3 = 0

. By testing values, we found

t = - 1

. If you were to analyze the derivative of

f (t) = t^{3} + 2 t + 3

, you would see

f^{'} (t) = 3 t^{2} + 2

, which is always positive. This means the function is strictly increasing and can only cross the x-axis once. Thus,

t = - 1

is the unique real root.

Why is the circle equation

(x - 2)^{2} + (y + 4)^{2} = 8

The circle is defined by its center and radius. We found the center

P

to be

(2, - 4)

. The circle passes through

C (0, - 6)

. The radius squared is the distance squared between

P

and

C

. Using the distance formula:

(2 - 0)^{2} + (- 4 - (- 6))^{2} = 2^{2} + 2^{2} = 4 + 4 = 8

. Hence,

(x - 2)^{2} + (y + 4)^{2} = 8

Animated Solution for Mathematics - Conic Sections: 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:

Select Answer:

Visualized Solution

Visualizing the Problem

The Minimum Distance Principle

Parametric Representation of $P$

Equation of the Normal

Normal Passing Through $C (0, - 6)$

Solving the Cubic Equation

Finding Coordinates of $P$

Calculating the Radius $R$

Equation of the New Circle

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Parametric Representation

Solving for the Parameter

Final Calculation

Similar Questions

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 $y^{2} = 4 x$ which is at the shortest distance from the center $S$ of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0$ . Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then

Let the shortest distance from $(a, 0)$ , $a > 0$ to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is:

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

The centre of the circle passing through the point $(0, 1)$ and touching the parabola $y = x^{2}$ at the point $(2, 4)$ is:

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , 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

Let a line perpendicular to the line $2 x - y = 10$ touch the parabola $y^{2} = 4 (x - 9)$ at the point $P$ . The distance of the point $P$ from the centre of the circle $x^{2} + y^{2} - 14 x - 8 y + 56 = 0$ is __________

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 radius of a circle, having minimum area, which touches the curve $y = 4 - x^{2}$ and the lines, $y = ∣ x ∣$ is:

Select Answer:

Visualized Solution

Visualizing the Problem

The Minimum Distance Principle

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

Equation of the Normal

Normal Passing Through .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C(0,−6)

Solving the Cubic Equation

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

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

Equation of the New Circle

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Parametric Representation

Solving for the Parameter

Final Calculation

Similar Questions

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The centre of the circle passing through the point (0,1) and touching the parabola y=x2 at the point (2,4) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola, y2=4x, 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 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 radius of a circle, having minimum area, which touches the curve y=4−x2 and the lines, y=∣x∣ is:

Parametric Representation of $P$

Normal Passing Through $C (0, - 6)$

Finding Coordinates of $P$

Calculating the Radius $R$

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

The centre of the circle passing through the point $(0, 1)$ and touching the parabola $y = x^{2}$ at the point $(2, 4)$ is:

Equation of a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , 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 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 radius of a circle, having minimum area, which touches the curve $y = 4 - x^{2}$ and the lines, $y = ∣ x ∣$ is: