JEE Advanced 1981

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Suppose that the normals drawn at three different points on the parabola $y^{2} = 4 x$ pass through the point $(h, k)$ . Show that $h > 2$ .

Visualized Solution

The Parabola $y^{2} = 4 x$

Consider the parabola $y^{2} = 4 x$ .
A point $(h, k)$ is given.
Three distinct normals are drawn from $(h, k)$ to the parabola.

Equation of Normal

The standard equation of a normal to $y^{2} = 4 a x$ in slope form is:
$y = m x - 2 am - a m^{3}$
Here, $m$ is the slope of the normal.

Normal for $y^{2} = 4 x$

Comparing $y^{2} = 4 x$ with $y^{2} = 4 a x$ , we get $4 a = 4 \Rightarrow a = 1$ .
Substituting $a = 1$ into the normal equation:
$y = m x - 2 m - m^{3}$

Passing Through $(h, k)$

The normal passes through the given point $(h, k)$ .
Substitute $x = h$ and $y = k$ :
$k = mh - 2 m - m^{3}$

The Cubic Equation in $m$

Rearrange the terms to form a standard cubic equation in $m$ :
$m^{3} + (2 - h) m + k = 0$
Or, $m^{3} - (h - 2) m + k = 0$

Three Distinct Normals

A cubic equation has at most three roots: $m_{1}, m_{2}, m_{3}$ .
Since three distinct normals can be drawn, the roots $m_{1}, m_{2}, m_{3}$ must be real and distinct.

Vieta's Formulas

For $m^{3} + 0 \cdot m^{2} - (h - 2) m + k = 0$ :
Sum of roots: $m_{1} + m_{2} + m_{3} = 0$
Sum of product of roots taken two at a time: $m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = - (h - 2)$

Algebraic Identity

We need to relate the roots to find a condition on $h$ .
Use the identity:
$(m_{1} + m_{2} + m_{3})^{2} = m_{1}^{2} + m_{2}^{2} + m_{3}^{2} + 2 (m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1})$

Substituting Vieta's Results

Substitute the values from Vieta's formulas into the identity:
$0^{2} = (m_{1}^{2} + m_{2}^{2} + m_{3}^{2}) + 2 [- (h - 2)]$

Sum of Squares of Slopes

Rearranging the equation to isolate the sum of squares:
$m_{1}^{2} + m_{2}^{2} + m_{3}^{2} = 2 (h - 2)$

Condition for Real Roots

The square of any non-zero real number is strictly positive.
Since $m_{1}, m_{2}, m_{3}$ are distinct real numbers (at most one can be zero), their sum of squares must be strictly greater than zero.
$m_{1}^{2} + m_{2}^{2} + m_{3}^{2} > 0$

Final Inequality

Substitute the expression for the sum of squares:
$2 (h - 2) > 0$
Dividing by 2:
$h - 2 > 0 \Rightarrow h > 2$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine standing on the coordinate plane, looking at the elegant curve of the parabola

y^{2} = 4 x

. It is a simple shape, yet it hides profound secrets.

We are given a point

(h, k)

and told that three distinct normals can be drawn from it to the parabola. Our mission is to prove that

h > 2

The Language of Normals

To understand the normals, we must speak their language. The equation of a normal to the parabola

y^{2} = 4 a x

with slope

m

is given by:

y = m x - 2 am - a m^{3}

By comparing our parabola

y^{2} = 4 x

to the standard form, we immediately identify

a = 1

. Substituting this, our normal equation becomes:

y = m x - 2 m - m^{3}

This equation is powerful because it encapsulates every possible normal to the parabola in terms of its slope

m

The Cubic Bridge

We are told the normal passes through

(h, k)

. This is the bridge between geometry and algebra. By substituting

x = h

and

y = k

into our normal equation, we get:

k = mh - 2 m - m^{3}

Rearranging this, we arrive at a cubic equation in

m

m^{3} + (2 - h) m + k = 0

The roots of this cubic,

m_{1}, m_{2}, m_{3}

, represent the slopes of the three distinct normals passing through

(h, k)

. The fact that there are three distinct normals means this cubic must have three distinct real roots.

The Vieta Insight

Now, we invoke the wisdom of Vieta. For our cubic

m^{3} + 0 \cdot m^{2} - (h - 2) m + k = 0

, the sum of the roots is

m_{1} + m_{2} + m_{3} = 0

The sum of the products of the roots taken two at a time is:

m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1} = - (h - 2)

We use the algebraic identity:

(m_{1} + m_{2} + m_{3})^{2} = m_{1}^{2} + m_{2}^{2} + m_{3}^{2} + 2 (m_{1} m_{2} + m_{2} m_{3} + m_{3} m_{1})

Substituting our Vieta relations, we get:

0^{2} = (m_{1}^{2} + m_{2}^{2} + m_{3}^{2}) + 2 [- (h - 2)]

This simplifies to:

m_{1}^{2} + m_{2}^{2} + m_{3}^{2} = 2 (h - 2)

The Final Revelation

Here is the moment of truth. We know that

m_{1}, m_{2}, m_{3}

are distinct real numbers.

The sum of the squares of distinct real numbers must be strictly greater than zero. Therefore:

2 (h - 2) > 0

This implies

h - 2 > 0

, or $h > 2$ .

We have arrived at the destination. The geometry of the parabola dictates that for three normals to exist, the point must lie to the right of

x = 2

Common Traps & FAQs

Why do we use the sum of squares of the slopes instead of the discriminant of the cubic?

The discriminant of a general cubic is a nightmare to calculate! Using the sum of squares is an elegant shortcut. Because we know the roots are real and distinct, their sum of squares must be strictly positive. It turns a complex calculus problem into a simple algebraic inequality.

Why does the problem specify 'distinct' normals?

Great catch! If the normals weren't distinct, the roots could be equal, which would allow the sum of squares to be zero. The word 'distinct' forces the inequality to be strict (

> 0

instead of

\geq 0

), which is exactly what we need to prove

h > 2

Is there a physical reason for

h > 2

Absolutely. The region where three normals can be drawn is bounded by the evolute of the parabola. For

y^{2} = 4 a x

, the evolute is

27 a y^{2} = 4 (x - 2 a)^{3}

. Our parabola has

a = 1

, so the evolute is

27 y^{2} = 4 (x - 2)^{3}

. Points to the right of

x = 2

are 'inside' the cusp of the evolute, allowing for three normals.

Animated Solution for Mathematics - Conic Sections: Suppose that the normals drawn at three different points on the parabola $y^{2} = 4 x$ pass through the point $(h, k)$ . Show that $h > 2$ .

Visualized Solution

The Parabola $y^{2} = 4 x$

Equation of Normal

Normal for $y^{2} = 4 x$

Passing Through $(h, k)$

The Cubic Equation in $m$

Three Distinct Normals

Vieta's Formulas

Algebraic Identity

Substituting Vieta's Results

Sum of Squares of Slopes

Condition for Real Roots

Final Inequality

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Language of Normals

The Cubic Bridge

The Vieta Insight

The Final Revelation

Similar Questions

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 three normals drawn to the parabola, $y^{2} = 2 x$ pass through the point $(a, 0) a \neq = 0$ , then 'a' must be greater than :

Three normals are drawn from the point $(c, 0)$ to the curve $y^{2} = x$ . Show that $c$ must be greater than $1/2$ . One normal is always the x-axis. Find $c$ for which the other two normals are perpendicular to each other.

$A$ is a point on the parabola $y^{2} = 4 a x$ . The normal at $A$ cuts the parabola again at point $B$ . If $A B$ subtends a right angle at the vertex of the parabola, find the slope of $A B$ .

Let $P (a sec θ, b tan θ)$ and $Q (a sec ϕ, b tan ϕ)$ , where $θ + ϕ = π /2$ , be two points on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$ , then $k$ is equal to

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

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

Consider the parabola with vertex $(\frac{1}{2}, \frac{3}{4})$ and the directrix $y = \frac{1}{2}$ . Let $P$ be the point where the parabola meets the line $x = - \frac{1}{2}$ . If the normal to the parabola at $P$ intersects the parabola again at the point $Q$ , then $(P Q)^{2}$ is equal to :

Match the following : $(3, 0)$ is the point from which three normals are drawn to the parabola $y^{2} = 4 x$ which meet the parabola in the points $P, Q$ and $R$ . Then

List-I

List-II

Visualized Solution

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

Equation of Normal

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

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

The Cubic Equation in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Three Distinct Normals

Vieta's Formulas

Algebraic Identity

Substituting Vieta's Results

Sum of Squares of Slopes

Condition for Real Roots

Final Inequality

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Language of Normals

The Cubic Bridge

The Vieta Insight

The Final Revelation

Similar Questions

.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 three normals drawn to the parabola, y2=2x pass through the point (a,0)a=0, then 'a' must be greater than :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A is a point on the parabola y2=4ax. The normal at A cuts the parabola again at point B. If AB subtends a right angle at the vertex of the parabola, find the slope of AB.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normals of the parabola y2=4x drawn at the end points of its latus rectum are tangents to the circle (x−3)2+(y+2)2=r2, then the value of r2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between a pair of tangents drawn from a point P to the parabola y2=4ax is 45∘. Show that the locus of the point P is a hyperbola.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Match the following : (3,0) is the point from which three normals are drawn to the parabola y2=4x which meet the parabola in the points P,Q and R. Then

List-I

List-II

The Parabola $y^{2} = 4 x$

Normal for $y^{2} = 4 x$

Passing Through $(h, k)$

The Cubic Equation in $m$

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 three normals drawn to the parabola, $y^{2} = 2 x$ pass through the point $(a, 0) a \neq = 0$ , then 'a' must be greater than :

$A$ is a point on the parabola $y^{2} = 4 a x$ . The normal at $A$ cuts the parabola again at point $B$ . If $A B$ subtends a right angle at the vertex of the parabola, find the slope of $A B$ .

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

The angle between a pair of tangents drawn from a point $P$ to the parabola $y^{2} = 4 a x$ is $4 5^{\circ}$ . Show that the locus of the point $P$ is a hyperbola.

Match the following : $(3, 0)$ is the point from which three normals are drawn to the parabola $y^{2} = 4 x$ which meet the parabola in the points $P, Q$ and $R$ . Then