JEE Advanced 1995

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Show that the locus of a point that divides a chord of slope 2 of the parabola $y^{2} = 4 x$ internally in the ratio $1 : 2$ is a parabola. Find the vertex of this parabola.

Q: Why do we use $t_1$ and $t_2$ instead of just solving for the intersection of the line and the parabola?

You *could* use $y=mx+c$, but you would end up with a quadratic equation in $x$ or $y$ with messy coefficients involving $c$. Parametric coordinates turn that mess into a clean, symmetric algebraic dance. It is not just easier; it is more elegant and less prone to calculation errors, which is vital in the JEE Advanced exam.

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the chord sliding along the parabola. Since the slope is fixed at 2, the chord is not just any line; it is a specific family of lines. The key is to realize that the endpoints

P

and

Q

are not independent—they are locked together by that slope constraint. Do not try to solve this with standard Cartesian lines (

y = m x + c

); it will lead to a swamp of quadratic roots.

Hint 2

Hint 2 (The Arsenal): Embrace the parametric form:

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

and

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

. The slope formula

m = \frac{2}{t _{1} + t _{2}}

is your bridge. By equating this to the given slope of 2, you will instantly reveal the hidden relationship

t_{1} + t_{2} = 1

. This is the 'golden key' that will simplify your entire algebraic journey.

Hint 3

Hint 3 (The Execution): Now, apply the section formula to find the coordinates of the point

R (h, k)

that divides

P Q

in the ratio

1 : 2

. You will have

h

and

k

expressed in terms of

t_{1}

and

t_{2}

. Use the constraint

t_{1} + t_{2} = 1

to eliminate

t_{2}

first, then express

t_{1}

in terms of

k

, and finally substitute everything into the expression for

h

. The resulting equation will be the locus.

Visualized Solution

Defining the Parabola and Points

Given Parabola: $y^{2} = 4 x$ (where $a = 1$ )
Let the endpoints of the chord be $P (t_{1}^{2}, 2 t_{1})$ and $Q (t_{2}^{2}, 2 t_{2})$

Slope of the Chord

Slope of chord $P Q$ : $m = \frac{2 t _{2} - 2 t _{1}}{t _{2}^{2} - t _{1}^{2}}$
Simplifying: $m = \frac{2 ( t _{2} - t _{1} )}{( t _{2} - t _{1} ) ( t _{2} + t _{1} )} = \frac{2}{t _{1} + t _{2}}$

Relation between $t_{1}$ and $t_{2}$

Given slope $m = 2$ : $\frac{2}{t _{1} + t _{2}} = 2$
Therefore, $t_{1} + t_{2} = 1 \Rightarrow t_{2} = 1 - t_{1}$

Applying the Section Formula

Let $R (h, k)$ divide $P Q$ in ratio $1 : 2$
$h = \frac{1 ( t _{2}^{2} ) + 2 ( t _{1}^{2} )}{3} = \frac{2 t _{1}^{2} + t _{2}^{2}}{3}$
$k = \frac{1 ( 2 t _{2} ) + 2 ( 2 t _{1} )}{3} = \frac{4 t _{1} + 2 t _{2}}{3}$

Expressing $k$ in terms of $t_{1}$

Substitute $t_{2} = 1 - t_{1}$ in $k$ :
$k = \frac{4 t _{1} + 2 ( 1 - t _{1} )}{3} = \frac{2 t _{1} + 2}{3}$
Solving for $t_{1}$ : $3 k = 2 t_{1} + 2 \Rightarrow t_{1} = \frac{3 k - 2}{2}$

Expressing $h$ in terms of $t_{1}$

Substitute $t_{2} = 1 - t_{1}$ in $h$ :
$3 h = 2 t_{1}^{2} + (1 - t_{1})^{2} = 2 t_{1}^{2} + 1 - 2 t_{1} + t_{1}^{2}$
$3 h = 3 t_{1}^{2} - 2 t_{1} + 1$

Eliminating the Parameter $t_{1}$

Substitute $t_{1} = \frac{3 k - 2}{2}$ into $3 h = 3 t_{1}^{2} - 2 t_{1} + 1$ :
$3 h = 3 (\frac{3 k - 2}{2})^{2} - 2 (\frac{3 k - 2}{2}) + 1$
$12 h = 3 (9 k^{2} - 12 k + 4) - 4 (3 k - 2) + 4$
$12 h = 27 k^{2} - 48 k + 24 \Rightarrow 4 h = 9 k^{2} - 16 k + 8$
Locus: $9 y^{2} - 16 y + 8 = 4 x$

Finding the Vertex

Rearranging: $9 (y^{2} - \frac{16}{9} y) = 4 x - 8$
Completing square: $9 (y - \frac{8}{9})^{2} = 4 x - 8 + 9 (\frac{64}{81})$
$9 (y - \frac{8}{9})^{2} = 4 x - \frac{8}{9} = 4 (x - \frac{2}{9})$
Vertex $(h, k) = (\frac{2}{9}, \frac{8}{9})$

Summary and Conclusion

Key Takeaway: The locus of a point dividing a chord of a parabola in a fixed ratio, given a fixed slope, is another parabola.
Final Answer: The vertex of the locus is $(\frac{2}{9}, \frac{8}{9})$ .

00:00 / 00:00

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

To find the locus of a point dividing a chord of the parabola

y^{2} = 4 x

in a specific ratio, we avoid the standard line equation

y = m x + c

to prevent complex quadratic roots. Instead, we utilize the parametric form.

We define the endpoints of the chord as

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

and

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

. This choice ensures that both points satisfy the equation of the parabola

y^{2} = 4 x

by construction.

The Slope Constraint

The slope

m

of the chord

P Q

is given by the change in

y

over the change in

x

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

Recognizing the denominator as a difference of squares, we factor it as

(t_{2} - t_{1}) (t_{2} + t_{1})

. Canceling the

(t_{2} - t_{1})

terms, we obtain:

m = \frac{2}{t _{1} + t _{2}}

Given that the slope is

2

, we set

\frac{2}{t _{1} + t _{2}} = 2

, which yields the fundamental constraint:

t_{1} + t_{2} = 1

The Section Formula

We seek the locus of a point

R (h, k)

that divides the chord

P Q

in the ratio

1 : 2

. Applying the section formula, we find:

h = \frac{1 ( t _{2}^{2} ) + 2 ( t _{1}^{2} )}{1 + 2} = \frac{t _{2}^{2} + 2 t _{1}^{2}}{3}

k = \frac{1 ( 2 t _{2} ) + 2 ( 2 t _{1} )}{1 + 2} = \frac{2 t _{2} + 4 t _{1}}{3}

Using the constraint

t_{2} = 1 - t_{1}

, we substitute into the expression for

k

k = \frac{2 ( 1 - t _{1} ) + 4 t _{1}}{3} = \frac{2 + 2 t _{1}}{3}

Solving for

t_{1}

, we find the parameter in terms of

k

t_{1} = \frac{3 k - 2}{2}

The Algebraic Dance

We now substitute

t_{1}

and

t_{2}

into the expression for

h

. Starting with

3 h = t_{2}^{2} + 2 t_{1}^{2}

and substituting

t_{2} = 1 - t_{1}

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

Substituting

t_{1} = \frac{3 k - 2}{2}

into this equation:

3 h = 3 (\frac{3 k - 2}{2})^{2} - 2 (\frac{3 k - 2}{2}) + 1

Expanding and simplifying leads to

12 h = 27 k^{2} - 48 k + 24

, which reduces to:

4 h = 9 k^{2} - 16 k + 8

Replacing

h

with

x

and

k

with

y

, the locus is:

9 y^{2} - 16 y + 8 = 4 x

Final Vertex Calculation

To identify the vertex, we complete the square for

y

9 (y^{2} - \frac{16}{9} y) = 4 x - 8

Adding

(\frac{8}{9})^{2}

inside the bracket (multiplied by 9) to both sides:

9 (y - \frac{8}{9})^{2} = 4 x - 8 + \frac{64}{9} = 4 x - \frac{8}{9} = 4 (x - \frac{2}{9})

The vertex of the resulting parabola is:

(\frac{2}{9}, \frac{8}{9})

Common Traps & FAQs

Why do we use

t_{1}

and

t_{2}

instead of just solving for the intersection of the line and the parabola?

You *could* use

y = m x + c

, but you would end up with a quadratic equation in

x

y

with messy coefficients involving

c

. Parametric coordinates turn that mess into a clean, symmetric algebraic dance. It is not just easier; it is more elegant and less prone to calculation errors, which is vital in the JEE Advanced exam.

How do I know the locus is a parabola?

Look at the final equation. If you have a variable squared (like

y^{2}

) and another variable linear (like

x

), that is the hallmark of a parabola. By eliminating the parameter

t_{1}

, we essentially projected the geometric constraint into a Cartesian equation. The degree of the equation remains quadratic in

y

and linear in

x

, which defines a parabola.

Is the ratio

1 : 2

arbitrary?

Not at all! The ratio determines the specific 'shape' and position of the resulting parabola. If the ratio were different, the vertex and the latus rectum of the new parabola would shift. The

1 : 2

ratio is the 'DNA' of this specific locus. Changing it would change the entire trajectory of the resulting curve.

Animated Solution for Mathematics - Conic Sections: Show that the locus of a point that divides a chord of slope 2 of the parabola $y^{2} = 4 x$ internally in the ratio $1 : 2$ is a parabola. Find the vertex of this parabola.

Visualized Solution

Defining the Parabola and Points

Slope of the Chord

Relation between $t_{1}$ and $t_{2}$

Applying the Section Formula

Expressing $k$ in terms of $t_{1}$

Expressing $h$ in terms of $t_{1}$

Eliminating the Parameter $t_{1}$

Finding the Vertex

Summary and Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Slope Constraint

The Section Formula

The Algebraic Dance

Final Vertex Calculation

Similar Questions

Let $O$ be the vertex of the parabola $x^{2} = 4 y$ and $Q$ be any point on it. Let the locus of the point $P$ , which divides the line segment $O Q$ internally in the ratio $2 : 3$ be the conic $C$ . Then the equation of the chord of $C$ , which is bisected at the point $(1, 2)$ , is :

Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2} = 4 x$ be the curve $S$ . Let $P$ be any point on $S$ . Then the locus of the point, which internally divides $O P$ in the ratio $3 : 1$ , is :

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, is :

Through the vertex $O$ of parabola $y^{2} = 4 x$ , chords $O P$ and $O Q$ are drawn at right angles to one another. Show that for all positions of $P$ , $P Q$ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of $P Q$ .

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is

Let $P$ be a variable point on the parabola $y = 4 x^{2} + 1$ . Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y = x$ is :

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $y^{2} = 4 a x$ is another parabola with directrix

The length of the chord of the parabola $x^{2} = 4 y$ having equation $x - 2 y + 42 = 0$ is :

If the chord joining the points $P_{1} (x_{1}, y_{1})$ and $P_{2} (x_{2}, y_{2})$ on the parabola $y^{2} = 12 x$ subtends a right angle at the vertex of the parabola, then $x_{1} x_{2} - y_{1} y_{2}$ is equal to

Visualized Solution

Defining the Parabola and Points

Slope of the Chord

Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2​

Applying the Section Formula

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​

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

Finding the Vertex

Summary and Conclusion

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Slope Constraint

The Section Formula

The Algebraic Dance

Final Vertex Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (x,y) be any point on the parabola y2=4x. Let P be the point that divides the line segment from (0,0) to (x,y) in the ratio 1:3. Then the locus of P is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y2=4ax is another parabola with directrix

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the chord of the parabola x2=4y having equation x−2​y+42​=0 is :

Relation between $t_{1}$ and $t_{2}$

Expressing $k$ in terms of $t_{1}$

Expressing $h$ in terms of $t_{1}$

Eliminating the Parameter $t_{1}$

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, is :

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $y^{2} = 4 a x$ is another parabola with directrix

The length of the chord of the parabola $x^{2} = 4 y$ having equation $x - 2 y + 42 = 0$ is :