JEE Main 2023 (08 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $R$ be the focus of the parabola $y^{2} = 20 x$ and the line $y = m x + c$ intersect the parabola at two points $P$ and $Q$ . Let the points $G (10, 10)$ be the centroid of the triangle $P QR$ . If $c - m = 6$ , then $P Q^{2}$ is

Q: Is the centroid formula really just the average of coordinates?

Yes, it is! For any triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the centroid is always $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$. It is a powerful tool that bridges the gap between geometry and algebra.

Select Answer:

Visualized Solution

Identify the Parabola and Focus

Given parabola: $y^{2} = 20 x$
Standard form: $y^{2} = 4 a x \Rightarrow 4 a = 20 \Rightarrow a = 5$
Focus $R = (a, 0) = (5, 0)$

Parametric Coordinates of $P$ and $Q$

Let $P$ and $Q$ be points on the parabola.
Parametric form: $(a t^{2}, 2 a t)$
$P = (5 t_{1}^{2}, 10 t_{1})$
$Q = (5 t_{2}^{2}, 10 t_{2})$

Centroid of $△ P QR$

Centroid $G (10, 10)$ of $△ P QR$
Formula: $G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$

Centroid $x$ -coordinate Setup

$x_{G} = \frac{x _{P} + x _{Q} + x _{R}}{3} = 10$
$\frac{5 t _{1}^{2} + 5 t _{2}^{2} + 5}{3} = 10$

Simplify for $t_{1}^{2} + t_{2}^{2}$

$5 (t_{1}^{2} + t_{2}^{2} + 1) = 30$
$t_{1}^{2} + t_{2}^{2} + 1 = 6$
$t_{1}^{2} + t_{2}^{2} = 5$

Centroid $y$ -coordinate Setup

$y_{G} = \frac{y _{P} + y _{Q} + y _{R}}{3} = 10$
$\frac{10 t _{1} + 10 t _{2} + 0}{3} = 10$

Simplify for $t_{1} + t_{2}$

$10 (t_{1} + t_{2}) = 30$
$t_{1} + t_{2} = 3$

Find the Product $t_{1} t_{2}$

Identity: $(t_{1} + t_{2})^{2} = t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2}$
Substitute knowns: $3^{2} = 5 + 2 t_{1} t_{2}$
$9 = 5 + 2 t_{1} t_{2} \Rightarrow 2 t_{1} t_{2} = 4$
$t_{1} t_{2} = 2$

Solve for Parameters $t_{1}$ and $t_{2}$

We have sum $t_{1} + t_{2} = 3$ and product $t_{1} t_{2} = 2$ .
They are roots of quadratic: $t^{2} - 3 t + 2 = 0$
$(t - 1) (t - 2) = 0 \Rightarrow t = 1, 2$
Let $t_{1} = 1$ and $t_{2} = 2$ .

Exact Coordinates of $P$ and $Q$

For $t_{1} = 1$ : $P = (5 (1)^{2}, 10 (1)) = (5, 10)$
For $t_{2} = 2$ : $Q = (5 (2)^{2}, 10 (2)) = (20, 20)$
Note: The condition $c - m = 6$ is consistent but not needed!

Calculate $P Q^{2}$

Distance formula squared: $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
$P Q^{2} = (20 - 5)^{2} + (20 - 10)^{2}$
$P Q^{2} = 1 5^{2} + 1 0^{2}$

Final Result

$P Q^{2} = 225 + 100$
$P Q^{2} = 325$
Final Answer: 325

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

Solution Diagram

Analyzing the Setup

The given parabola is defined by the equation

y^{2} = 20 x

. By comparing this to the standard form

y^{2} = 4 a x

, we identify

4 a = 20

, which yields

a = 5

Consequently, the focus

R

of the parabola is located at the coordinates

(5, 0)

The Parametric Approach

To handle the intersection points

P

and

Q

efficiently, we utilize parametric coordinates. For a parabola

y^{2} = 4 a x

, any point can be represented as

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

With

a = 5

, we define the points as:

P = (5 t_{1}^{2}, 10 t_{1})

Q = (5 t_{2}^{2}, 10 t_{2})

Applying the Centroid Condition

The centroid

G

△ P QR

is given as

(10, 10)

. The formula for the centroid of a triangle with vertices

(x_{1}, y_{1})

(x_{2}, y_{2})

, and

(x_{3}, y_{3})

is:

G = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})

Substituting the coordinates of

P

Q

, and

R (5, 0)

into the centroid formula, we obtain two equations:

For the x-coordinate:

\frac{5 t _{1}^{2} + 5 t _{2}^{2} + 5}{3} = 10 \Rightarrow t_{1}^{2} + t_{2}^{2} = 5

For the y-coordinate:

\frac{10 t _{1} + 10 t _{2} + 0}{3} = 10 \Rightarrow t_{1} + t_{2} = 3

Solving for Parameters

We use the algebraic identity

(t_{1} + t_{2})^{2} = t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2}

to find the product of the parameters. Substituting our known values:

3^{2} = 5 + 2 t_{1} t_{2}

9 = 5 + 2 t_{1} t_{2} \Rightarrow t_{1} t_{2} = 2

The values

t_{1}

and

t_{2}

are the roots of the quadratic equation

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

. Solving this, we find

t = 1

and

t = 2

Final Calculation

Using these parameters, we determine the coordinates of the points:

P = (5 (1)^{2}, 10 (1)) = (5, 10)

Q = (5 (2)^{2}, 10 (2)) = (20, 20)

Finally, we calculate the square of the distance

P Q

P Q^{2} = (20 - 5)^{2} + (20 - 10)^{2}

P Q^{2} = 1 5^{2} + 1 0^{2} = 225 + 100

The final result is $P Q^{2} = 325$ .

Common Traps & FAQs

Why did we ignore the line equation

y = m x + c

Great question! In many JEE problems, you are given extra information to test if you can identify the necessary constraints. Since the centroid

G

and the focus

R

are fixed, the points

P

and

Q

are already uniquely determined. The line equation is consistent with these points, but it is not required to solve for the distance

P Q

How do I know to use parametric coordinates instead of the standard

y = m x + c

approach?

Excellent intuition. While you could solve this using the intersection of the line and parabola, it leads to a messy quadratic in

x

. Parametric coordinates

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

are the 'gold standard' for parabola problems because they reduce the degree of the equations and make the algebra symmetric and elegant.

Is the centroid formula really just the average of coordinates?

Yes, it is! For any triangle with vertices

(x_{1}, y_{1})

(x_{2}, y_{2})

, and

(x_{3}, y_{3})

, the centroid is always

(\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})

. It is a powerful tool that bridges the gap between geometry and algebra.

Animated Solution for Mathematics - Conic Sections: Let $R$ be the focus of the parabola $y^{2} = 20 x$ and the line $y = m x + c$ intersect the parabola at two points $P$ and $Q$ . Let the points $G (10, 10)$ be the centroid of the triangle $P QR$ . If $c - m = 6$ , then $P Q^{2}$ is

Select Answer:

Visualized Solution

Identify the Parabola and Focus

Parametric Coordinates of $P$ and $Q$

Centroid of $△ P QR$

Centroid $x$ -coordinate Setup

Simplify for $t_{1}^{2} + t_{2}^{2}$

Centroid $y$ -coordinate Setup

Simplify for $t_{1} + t_{2}$

Find the Product $t_{1} t_{2}$

Solve for Parameters $t_{1}$ and $t_{2}$

Exact Coordinates of $P$ and $Q$

Calculate $P Q^{2}$

Final Result

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

Analyzing the Setup

The Parametric Approach

Applying the Centroid Condition

Solving for Parameters

Final Calculation

Similar Questions

Let $A$ be the focus of the parabola $y^{2} = 8 x$ . Let the line $y = m x + c$ intersect the parabola at two distinct points $B$ and $C$ . If the centroid of the triangle $A B C$ is $(\frac{7}{3}, \frac{4}{3})$ , then $(B C)^{2}$ is equal to :

Let $y^{2} = 12 x$ be the parabola with its vertex at $O$ . Let $P$ be a point on the parabola and $A$ be a point on the $x$ -axis such that $∠ O P A = 9 0^{\circ}$ . Then the locus of the centroid of such triangles $O P A$ is :

Let $A (0, 1)$ , $B (1, 1)$ and $C (1, 0)$ be the mid-points of the sides of a triangle with incentre at the point $D$ . If the focus of the parabola $y^{2} = 4 a x$ passing through $D$ is $(α + β 2, 0)$ , where $α$ and $β$ are rational numbers, then $\frac{α}{β ^{2}}$ is equal to

Let the point $P$ of the focal chord $P Q$ of the parabola $y^{2} = 16 x$ be $(1, - 4)$ . If the focus of the parabola divides the chord $P Q$ in the ratio $m : n$ , $gcd (m, n) = 1$ , then $m^{2} + n^{2}$ is equal to :

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , and its foci is :

Let $P (4, 43)$ be a point on the parabola $y^{2} = 4 a x$ and PQ be a focal chord of the parabola. If M and N are the foot of perpendiculars drawn from P and Q respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to:

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ 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

Select Answer:

Visualized Solution

Identify the Parabola and Focus

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

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Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate Setup

Simplify for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t12​+t22​

Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate Setup

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

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

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Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ2

Final Result

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

Analyzing the Setup

The Parametric Approach

Applying the Centroid Condition

Solving for Parameters

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2−9y2+32x+36y−164=0, and its foci is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the focus of the parabola y2=8x and let PQ be the common chord of the circle x2+y2−2x−4y=0 and the given parabola. The area of the triangle PQS is

Parametric Coordinates of $P$ and $Q$

Centroid of $△ P QR$

Centroid $x$ -coordinate Setup

Simplify for $t_{1}^{2} + t_{2}^{2}$

Centroid $y$ -coordinate Setup

Simplify for $t_{1} + t_{2}$

Find the Product $t_{1} t_{2}$

Solve for Parameters $t_{1}$ and $t_{2}$

Exact Coordinates of $P$ and $Q$

Calculate $P Q^{2}$

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , and its foci is :

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is