JEE Advanced 2015

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Suppose that the foci of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ are $(f_{1}, 0)$ and $(f_{2}, 0)$ where $f_{1} > 0$ and $f_{2} < 0$ . Let $P_{1}$ and $P_{2}$ be two parabolas with a common vertex at $(0, 0)$ and with foci at $(f_{1}, 0)$ and $(2 f_{2}, 0)$ , respectively. Let $T_{1}$ be a tangent to $P_{1}$ which passes through $(2 f_{2}, 0)$ and $T_{2}$ be a tangent to $P_{2}$ which passes through $(f_{1}, 0)$ . If $m_{1}$ is the slope of $T_{1}$ and $m_{2}$ is the slope of $T_{2}$ , then the value of $(\frac{1}{m _{1}^{2}} + m_{2}^{2})$ is

Q: Wait, why is the equation for $P_2$ $y^2 = -16x$? Shouldn't it be $y^2 = 16x$?

Great catch! Remember, the parabola's orientation depends on the focus. Since $f_2 < 0$, the focus is at $(-4, 0)$. A parabola with a focus on the negative x-axis must open to the left. Thus, $4a = -16$, making the equation $y^2 = -16x$.

Q: Why do we use the slope form $y = mx + a/m$ instead of the point-slope form?

You *could* use the point-slope form, but it would require solving a quadratic equation in $m$ by setting the discriminant to zero. The slope form is a shortcut derived from that exact process. In JEE Advanced, efficiency is key!

Q: Why don't we need to find the actual values of $m_1$ and $m_2$?

Look at the final expression: $(\frac{1}{m_1^2} + m_2^2)$. The question is clever! It only asks for the squares of the slopes. Solving for $m_1$ or $m_2$ individually would involve square roots and unnecessary complexity. Always keep your eyes on the target expression.

Enter Numerical Value:

Progressive Hints

Hint 1

Visualize the geometry first. The ellipse is the 'master' curve here. Its foci,

(f_{1}, 0)

and

(f_{2}, 0)

, are the anchors for your parabolas. Don't jump into algebra until you have plotted these points on the Cartesian plane. Where do they lie relative to the origin?

Hint 2

You need the 'Slope Form' of the tangent to a parabola. For a parabola

y^{2} = 4 a x

, the tangent with slope

m

is given by

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

. This is your most powerful tool. It connects the geometry of the tangent line directly to the algebraic slope

m

Hint 3

Now, execute the condition of tangency. For

P_{1}

, substitute the point

(2 f_{2}, 0)

into the tangent equation

y = m_{1} x + \frac{2}{m _{1}}

. For

P_{2}

, substitute

(f_{1}, 0)

into

y = m_{2} x - \frac{4}{m _{2}}

. This will give you two simple algebraic equations for

m_{1}^{2}

and

m_{2}^{2}

Visualized Solution

$\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$

Ellipse Equation: $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$
Identify parameters: $a^{2} = 9 \Rightarrow a = 3$ and $b^{2} = 5 \Rightarrow b = 5$

$e = 1 - \frac{b ^{2}}{a ^{2}}$

Calculate Eccentricity: $e = 1 - \frac{b ^{2}}{a ^{2}}$
$e = 1 - \frac{5}{9} = \frac{4}{9} = \frac{2}{3}$

$(\pm a e, 0)$

Foci coordinates: $(\pm a e, 0) = (\pm 3 \cdot \frac{2}{3}, 0) = (\pm 2, 0)$
Given $f_{1} > 0$ , we have $f_{1} = 2$
Given $f_{2} < 0$ , we have $f_{2} = - 2$

$P_{1} : y^{2} = 8 x$

Parabola $P_{1}$ : Vertex $(0, 0)$ , Focus $(f_{1}, 0) = (2, 0)$
Standard form $y^{2} = 4 a x$ with $a = 2$
Equation of $P_{1}$ : $y^{2} = 4 (2) x \Rightarrow y^{2} = 8 x$

$T_{1} : y = m_{1} x + \frac{2}{m _{1}}$

Tangent $T_{1}$ to $y^{2} = 4 a x$ has slope form: $y = m x + \frac{a}{m}$
For $P_{1}$ , $a = 2$ and slope is $m_{1}$
Equation of $T_{1}$ : $y = m_{1} x + \frac{2}{m _{1}}$

Substitute $(2 f_{2}, 0)$

Condition: $T_{1}$ passes through $(2 f_{2}, 0)$
Calculate point: $(2 (- 2), 0) = (- 4, 0)$
Substitute $(- 4, 0)$ into $T_{1}$ : $0 = m_{1} (- 4) + \frac{2}{m _{1}}$

$m_{1}^{2} = \frac{1}{2}$

Rearrange equation: $4 m_{1} = \frac{2}{m _{1}}$
Cross-multiply: $4 m_{1}^{2} = 2$
Result: $m_{1}^{2} = \frac{1}{2}$

$P_{2} : y^{2} = - 16 x$

Parabola $P_{2}$ : Vertex $(0, 0)$ , Focus $(2 f_{2}, 0) = (- 4, 0)$
Standard form $y^{2} = 4 a x$ with $a = - 4$
Equation of $P_{2}$ : $y^{2} = 4 (- 4) x \Rightarrow y^{2} = - 16 x$

$T_{2} : y = m_{2} x - \frac{4}{m _{2}}$

For $P_{2}$ , $a = - 4$ and slope is $m_{2}$
Equation of $T_{2}$ : $y = m_{2} x + \frac{- 4}{m _{2}}$
Simplified: $y = m_{2} x - \frac{4}{m _{2}}$

Substitute $(f_{1}, 0)$

Condition: $T_{2}$ passes through $(f_{1}, 0) = (2, 0)$
Substitute $(2, 0)$ into $T_{2}$ : $0 = m_{2} (2) - \frac{4}{m _{2}}$

$m_{2}^{2} = 2$

Rearrange equation: $2 m_{2} = \frac{4}{m _{2}}$
Cross-multiply: $2 m_{2}^{2} = 4$
Result: $m_{2}^{2} = 2$

$\frac{1}{m _{1}^{2}} + m_{2}^{2}$

Target Expression: $\frac{1}{m _{1}^{2}} + m_{2}^{2}$
Substitute $m_{1}^{2} = \frac{1}{2}$ and $m_{2}^{2} = 2$ :
$\frac{1}{1/2} + 2 = 2 + 2 = 4$
Final Answer: 4

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

Solution Diagram

Analyzing the Setup

The Conic Dance: A Journey Through Geometry and Algebra. Welcome, future engineer! Today, we are not just solving a problem; we are choreographing a dance between three distinct conic sections. We have an ellipse, the elegant parent, and two parabolas, its energetic offspring.

Unveiling the Ellipse

We start with the equation:

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

By comparing this to the standard form

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

, we immediately see that

a^{2} = 9

and

b^{2} = 5

. This tells us that

a = 3

and

b = 5

To find the foci, we calculate the eccentricity

e

e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{5}{9} = \frac{4}{9} = \frac{2}{3}

The foci are located at

(\pm a e, 0)

. Substituting our values, we get

(\pm 3 \cdot \frac{2}{3}, 0)

, which simplifies to

(\pm 2, 0)

. Thus, our anchors are

f_{1} = 2

and

f_{2} = - 2

Building the Parabolas

Now, we construct our parabolas.

P_{1}

has its vertex at the origin and its focus at

(f_{1}, 0) = (2, 0)

. Using the form

y^{2} = 4 a x

with

a = 2

, we get:

P_{1} : y^{2} = 8 x

Next,

P_{2}

has its vertex at the origin and its focus at

(2 f_{2}, 0) = (2 (- 2), 0) = (- 4, 0)

. With

a = - 4

, the equation becomes:

P_{2} : y^{2} = - 16 x

The Tangent Bridge

For any parabola

y^{2} = 4 a x

, the tangent with slope

m

is given by

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

. This is the bridge between the geometry of the curve and the algebra of the line.

For

P_{1}

, where

a = 2

, the tangent

T_{1}

y = m_{1} x + \frac{2}{m _{1}}

. Since

T_{1}

passes through

(- 4, 0)

, we substitute:

0 = m_{1} (- 4) + \frac{2}{m _{1}} \Rightarrow 4 m_{1} = \frac{2}{m _{1}} \Rightarrow m_{1}^{2} = \frac{1}{2}

For

P_{2}

, where

a = - 4

, the tangent

T_{2}

y = m_{2} x - \frac{4}{m _{2}}

. Since

T_{2}

passes through

(2, 0)

, we substitute:

0 = m_{2} (2) - \frac{4}{m _{2}} \Rightarrow 2 m_{2} = \frac{4}{m _{2}} \Rightarrow m_{2}^{2} = 2

Final Calculation

We have arrived at the final step. The problem asks for the value of

(\frac{1}{m _{1}^{2}} + m_{2}^{2})

Substituting our derived values

m_{1}^{2} = \frac{1}{2}

and

m_{2}^{2} = 2

\frac{1}{1/2} + 2 = 2 + 2 = 4

The final result is 4. You have mastered the geometry; keep this confidence and carry it into your next challenge!

Common Traps & FAQs

Wait, why is the equation for

P_{2}

y^{2} = - 16 x

? Shouldn't it be

y^{2} = 16 x

Great catch! Remember, the parabola's orientation depends on the focus. Since

f_{2} < 0

, the focus is at

(- 4, 0)

. A parabola with a focus on the negative x-axis must open to the left. Thus,

4 a = - 16

, making the equation

y^{2} = - 16 x

Why do we use the slope form

y = m x + a / m

instead of the point-slope form?

You *could* use the point-slope form, but it would require solving a quadratic equation in

m

by setting the discriminant to zero. The slope form is a shortcut derived from that exact process. In JEE Advanced, efficiency is key!

Why don't we need to find the actual values of

m_{1}

and

m_{2}

Look at the final expression:

(\frac{1}{m _{1}^{2}} + m_{2}^{2})

. The question is clever! It only asks for the squares of the slopes. Solving for

m_{1}

m_{2}

individually would involve square roots and unnecessary complexity. Always keep your eyes on the target expression.

Enter Numerical Value:

Visualized Solution

$\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$

$e = 1 - \frac{b ^{2}}{a ^{2}}$

$(\pm a e, 0)$

$P_{1} : y^{2} = 8 x$

$T_{1} : y = m_{1} x + \frac{2}{m _{1}}$

Substitute $(2 f_{2}, 0)$

$m_{1}^{2} = \frac{1}{2}$

$P_{2} : y^{2} = - 16 x$

$T_{2} : y = m_{2} x - \frac{4}{m _{2}}$

Substitute $(f_{1}, 0)$

$m_{2}^{2} = 2$

$\frac{1}{m _{1}^{2}} + m_{2}^{2}$

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

Analyzing the Setup

Unveiling the Ellipse

Building the Parabolas

The Tangent Bridge

Final Calculation

Similar Questions

Comprehension Passage

The orthocentre of the triangle $F_{1} M N$ is

If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the x-axis at $Q$ , then the ratio of area of the triangle $M QR$ to area of the quadrilateral $M F_{1} N F_{2}$ is

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P (\frac{1}{2}, 1)$ . Its one directrix is the common tangent, nearer to the point $P$ , to the circle $x^{2} + y^{2} = 1$ and the hyperbola $x^{2} - y^{2} = 1$ . The equation of the ellipse, in the standard form, is.........

The foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{81} = \frac{1}{25}$ coincide. Then the value of $b^{2}$ is

Enter Numerical Value:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e=1−a2b2​​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​:y2=8x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T1​:y=m1​x+m1​2​

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2f2​,0)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m12​=21​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​:y2=−16x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​:y=m2​x−m2​4​

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (f1​,0)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m22​=2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m12​1​+m22​

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

Analyzing the Setup

Unveiling the Ellipse

Building the Parabolas

The Tangent Bridge

Final Calculation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The orthocentre of the triangle F1​MN is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The foci of the ellipse 16x2​+b2y2​=1 and the hyperbola 144x2​−81y2​=251​ coincide. Then the value of b2 is

$\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$

$e = 1 - \frac{b ^{2}}{a ^{2}}$

$(\pm a e, 0)$

$P_{1} : y^{2} = 8 x$

$T_{1} : y = m_{1} x + \frac{2}{m _{1}}$

Substitute $(2 f_{2}, 0)$

$m_{1}^{2} = \frac{1}{2}$

$P_{2} : y^{2} = - 16 x$

$T_{2} : y = m_{2} x - \frac{4}{m _{2}}$

Substitute $(f_{1}, 0)$

$m_{2}^{2} = 2$

$\frac{1}{m _{1}^{2}} + m_{2}^{2}$

The orthocentre of the triangle $F_{1} M N$ is

The foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{81} = \frac{1}{25}$ coincide. Then the value of $b^{2}$ is