JEE Advanced 2014

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: The common tangents to the circle $x^{2} + y^{2} = 2$ and the parabola $y^{2} = 8 x$ touch the circle at the points $P, Q$ and the parabola at the points $R, S$ . Then the area of the quadrilateral $P QR S$ is

Q: How do I know if the tangents are real?

When you solve the equation for $m$, you will get a biquadratic equation. If the roots for $m^2$ are positive, the slopes are real. If they were negative, it would imply no real common tangents exist. In this problem, $m^2 = 1$ gives us $m = \pm 1$, confirming that two real common tangents exist.

Select Answer:

Visualized Solution

Visualizing the Curves

Given Circle: $x^{2} + y^{2} = 2$ (Center $(0, 0)$ , Radius $r = 2$ )
Given Parabola: $y^{2} = 8 x$ (Comparing with $y^{2} = 4 a x$ , we get $a = 2$ )

General Tangent to the Parabola

The general equation of a tangent to the parabola $y^{2} = 4 a x$ is $y = m x + \frac{a}{m}$ .
For $y^{2} = 8 x$ , where $a = 2$ , the tangent is: $y = m x + \frac{2}{m}$
Rearranging to standard form: $m x - y + \frac{2}{m} = 0$

Condition for Tangency to the Circle

A line $A x + B y + C = 0$ touches a circle $x^{2} + y^{2} = r^{2}$ if the perpendicular distance from the center equals the radius: $\frac{∣ C ∣}{A ^{2} + B ^{2}} = r$ .
Applying this to $m x - y + \frac{2}{m} = 0$ and $r = 2$ :
$\frac{∣ \frac{2}{m} ∣}{m ^{2} + ( - 1 ) ^{2}} = 2$

Solving for the Slope $m$

Square both sides: $\frac{4}{m ^{2} ( m ^{2} + 1 )} = 2$
Cross-multiply: $2 = m^{2} (m^{2} + 1) ⟹ m^{4} + m^{2} - 2 = 0$
Factorize: $(m^{2} + 2) (m^{2} - 1) = 0$
Since $m^{2} \neq = - 2$ (for real slopes), we have $m^{2} = 1 ⟹ m = \pm 1$

The Common Tangent Equations

Substitute $m = 1$ into $y = m x + \frac{2}{m}$ : $y = x + 2$
Substitute $m = - 1$ into $y = m x + \frac{2}{m}$ : $y = - x - 2$
These are the two common tangents to the circle and the parabola.

Points of Contact on the Circle

The point of contact of a tangent $y = m x \pm r 1 + m^{2}$ to $x^{2} + y^{2} = r^{2}$ is $(\frac{\mp m r}{1 + m ^{2}}, \frac{\pm r}{1 + m ^{2}})$ .
Alternatively, solve $y = x + 2$ with $x^{2} + y^{2} = 2$ : $x^{2} + (x + 2)^{2} = 2 ⟹ 2 x^{2} + 4 x + 2 = 0 ⟹ (x + 1)^{2} = 0 ⟹ x = - 1, y = 1$ . So, $P (- 1, 1)$ .
Similarly, for $y = - x - 2$ : $Q (- 1, - 1)$ .

Points of Contact on the Parabola

The point of contact of tangent $y = m x + \frac{a}{m}$ to $y^{2} = 4 a x$ is given by $(\frac{a}{m ^{2}}, \frac{2 a}{m})$ .
For $m = 1, a = 2$ : $R = (\frac{2}{1 ^{2}}, \frac{2 ( 2 )}{1}) = (2, 4)$ .
For $m = - 1, a = 2$ : $S = (\frac{2}{( - 1 ) ^{2}}, \frac{2 ( 2 )}{- 1}) = (2, - 4)$ .

Identifying the Quadrilateral $P QR S$

Vertices: $P (- 1, 1), Q (- 1, - 1), R (2, 4), S (2, - 4)$
Notice that $P$ and $Q$ have the same x-coordinate ( $x = - 1$ ). So, line segment $P Q$ is vertical.
Similarly, $R$ and $S$ have the same x-coordinate ( $x = 2$ ). So, line segment $R S$ is also vertical.
Since $P Q$ and $R S$ are both vertical, they are parallel to each other.
Therefore, the quadrilateral $P QR S$ is a trapezium.

Calculating the Area

Length of parallel side $P Q = 1 - (- 1) = 2$
Length of parallel side $R S = 4 - (- 4) = 8$
The perpendicular distance (height $h$ ) between the parallel sides $x = - 1$ and $x = 2$ is $h = 2 - (- 1) = 3$ .
Area of Trapezium $= \frac{1}{2} \times (Sum of parallel sides) \times height$
Area $= \frac{1}{2} \times (2 + 8) \times 3 = \frac{1}{2} \times 10 \times 3 = 15$

Conclusion and Key Takeaway

Final Answer: The area of quadrilateral $P QR S$ is $15$ square units.
Key Takeaway: Using the parametric point of contact $(\frac{a}{m ^{2}}, \frac{2 a}{m})$ for a parabola significantly speeds up calculations compared to solving equations simultaneously.
Pro Tip: Always look for geometric properties like vertical or horizontal lines to simplify area calculations instead of blindly using the shoelace formula.

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

Solution Diagram

Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane. Before you lies a circle,

x^{2} + y^{2} = 2

, a perfect, symmetric loop with a radius of

2

To your right, the parabola

y^{2} = 8 x

opens its arms wide. Your mission is to find the common tangents—the lines that graze both these curves simultaneously.

The Tangent Hunt

To bridge these two curves, we need a line that speaks the language of both. For the parabola

y^{2} = 4 a x

, where

a = 2

, any tangent can be described by the elegant equation:

y = m x + \frac{2}{m}

This equation is our bridge. It doesn't matter where the tangent touches the parabola; as long as it has slope

m

, it must satisfy this form.

Now, we bring in the circle. For this line to touch the circle, the perpendicular distance from the center

(0, 0)

to the line

m x - y + \frac{2}{m} = 0

must be exactly the radius,

2

Using the distance formula, we get:

\frac{∣ \frac{2}{m} ∣}{m ^{2} + 1} = 2

Squaring both sides, we arrive at:

\frac{4}{m ^{2} ( m ^{2} + 1 )} = 2

This simplifies to the beautiful biquadratic equation:

m^{4} + m^{2} - 2 = 0

Factoring this, we find

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

. Since

m^{2}

cannot be

- 2

for real slopes, we are left with

m^{2} = 1

, giving us

m = \pm 1

The common tangents are $y = x + 2$ and $y = - x - 2$ .

The Geometry of Contact

Now that we have the lines, we need the points where they touch the curves. For the circle, solving

y = x + 2

with

x^{2} + y^{2} = 2

leads us to

x^{2} + (x + 2)^{2} = 2

, which simplifies to

2 (x + 1)^{2} = 0

This tells us the tangent touches the circle at

x = - 1

. Substituting back, we get

y = 1

. So, the points of contact on the circle are

P (- 1, 1)

and, by symmetry,

Q (- 1, - 1)

For the parabola, we use the parametric point of contact

(\frac{a}{m ^{2}}, \frac{2 a}{m})

. With

a = 2

and

m = 1

, we get

R (2, 4)

. With

m = - 1

, we get

S (2, - 4)

The Trapezium Insight

Look at the coordinates:

P (- 1, 1)

Q (- 1, - 1)

R (2, 4)

, and

S (2, - 4)

. Notice that

P

and

Q

share the same

x

-coordinate, making

P Q

a vertical line of length

1 - (- 1) = 2

Similarly,

R

and

S

share an

x

-coordinate, making

R S

a vertical line of length

4 - (- 4) = 8

. Because both

P Q

and

R S

are vertical, they are parallel.

We have formed a trapezium. The distance between these parallel lines is the difference in their

x

-coordinates:

2 - (- 1) = 3

The area is calculated as:

Area = \frac{1}{2} \times (2 + 8) \times 3 = 15

We have conquered the problem not by brute force, but by observing the geometric soul of the equations. The final area is 15.

Common Traps & FAQs

Wait, why do we use

y = m x + \frac{a}{m}

instead of the point-slope form?

Great question! The point-slope form requires you to know the point of contact beforehand, which we don't. The form

y = m x + \frac{a}{m}

is the 'slope form' of the tangent. It is the most efficient tool here because it directly relates the slope

m

to the parabola's parameter

a

, allowing us to use the circle's tangency condition to find

m

immediately.

Why can't we just use the shoelace formula for the area?

You absolutely could! The shoelace formula is a universal tool for any polygon. However, in JEE Advanced, time is your most precious resource. By identifying that the quadrilateral is a trapezium (because the sides

P Q

and

R S

are vertical), we can use the much simpler formula

\frac{1}{2} \times (sum of parallel sides) \times height

. It's about working smarter, not just harder.

How do I know if the tangents are real?

When you solve the equation for

m

, you will get a biquadratic equation. If the roots for

m^{2}

are positive, the slopes are real. If they were negative, it would imply no real common tangents exist. In this problem,

m^{2} = 1

gives us

m = \pm 1

, confirming that two real common tangents exist.

Animated Solution for Mathematics - Conic Sections: The common tangents to the circle $x^{2} + y^{2} = 2$ and the parabola $y^{2} = 8 x$ touch the circle at the points $P, Q$ and the parabola at the points $R, S$ . Then the area of the quadrilateral $P QR S$ is

Select Answer:

Visualized Solution

Visualizing the Curves

General Tangent to the Parabola

Condition for Tangency to the Circle

Solving for the Slope $m$

The Common Tangent Equations

Points of Contact on the Circle

Points of Contact on the Parabola

Identifying the Quadrilateral $P QR S$

Calculating the Area

Conclusion and Key Takeaway

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

Analyzing the Setup

The Tangent Hunt

The Geometry of Contact

The Trapezium Insight

Similar Questions

Comprehension Passage

The ratio of the areas of the triangles $P QS$ and $P QR$ is

The radius of the circumcircle of the triangle $P R S$ is

The radius of the incircle of the triangle $P QR$ is

From a point $A$ common tangents are drawn to the circle $x^{2} + y^{2} = a^{2} /2$ and parabola $y^{2} = 4 a x$ . Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola.

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

Two common tangents to the circle $x^{2} + y^{2} = 2 a^{2}$ and parabola $y^{2} = 8 a x$ are

Tangents drawn from the point (-8, 0) to the parabola $y^{2} = 8 x$ touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :-

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:

If the points of intersection of two distinct conics $x^{2} + y^{2} = 4 b$ and $\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1$ lie on the curve $y^{2} = 3 x^{2}$ , then $33$ times the area of the rectangle formed by the intersection points is ______

A line is a common tangent to the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ . If the two points of contact $(a, b)$ and $(c, d)$ are distinct and lie in the first quadrant, then $2 (a + c)$ is equal to

Select Answer:

Visualized Solution

Visualizing the Curves

General Tangent to the Parabola

Condition for Tangency to the Circle

Solving for the Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

The Common Tangent Equations

Points of Contact on the Circle

Points of Contact on the Parabola

Identifying the Quadrilateral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQRS

Calculating the Area

Conclusion and Key Takeaway

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

Analyzing the Setup

The Tangent Hunt

The Geometry of Contact

The Trapezium Insight

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The ratio of the areas of the triangles PQS and PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the circumcircle of the triangle PRS is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the incircle of the triangle PQR 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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two common tangents to the circle x2+y2=2a2 and parabola y2=8ax are

Solving for the Slope $m$

Identifying the Quadrilateral $P QR S$

The ratio of the areas of the triangles $P QS$ and $P QR$ is

The radius of the circumcircle of the triangle $P R S$ is

The radius of the incircle of the triangle $P QR$ 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

Two common tangents to the circle $x^{2} + y^{2} = 2 a^{2}$ and parabola $y^{2} = 8 a x$ are