JEE Advanced 2015

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let $E_{1}$ and $E_{2}$ be two ellipses whose centers are at the origin. The major axes of $E_{1}$ and $E_{2}$ lie along the $x$ -axis and the $y$ -axis, respectively. Let $S$ be the circle $x^{2} + (y - 1)^{2} = 2$ . The straight line $x + y = 3$ touches the curves $S, E_{1}$ and $E_{2}$ at $P, Q$ and $R$ respectively. Suppose that $P Q = P R = \frac{2 2}{3}$ . If $e_{1}$ and $e_{2}$ are the eccentricities of $E_{1}$ and $E_{2}$ , respectively, then the correct expression(s) is (are)

Q: How do I know which point is $Q$ and which is $R$?

Great question! The problem states $E_1$ has its major axis along the $x$-axis and $E_2$ along the $y$-axis. When you calculate the two points using the parametric distance, you will get two sets of coordinates. One set will yield $a_1^2 > b_1^2$ (consistent with $E_1$), and the other will yield $a_2^2 > b_2^2$ (consistent with $E_2$). The geometry naturally sorts itself out if you follow the constraints of the major axes.

Q: Why is the slope of the normal $1$?

Remember that the tangent line is $x + y = 3$, which can be rewritten as $y = -x + 3$. The slope of this line is $-1$. A normal is, by definition, perpendicular to the tangent at the point of contact. If two lines are perpendicular, the product of their slopes is $-1$. So, if the tangent slope is $-1$, the normal slope must be $1$ because $(-1) \times (1) = -1$.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

The Spark: Visualize the geometry. You have a circle and two ellipses sharing a common tangent line. The circle is the anchor. The normal to the circle at the point of contact

P

must pass through the circle's center. Since the tangent line

x + y = 3

has a slope of

- 1

, the normal must have a slope of

1

. Use this to find

P

Hint 2

The Arsenal: Once you have

P (1, 2)

, you need to find

Q

and

R

. The line

x + y = 3

has an inclination of

13 5^{\circ}

. Use the parametric form of a line:

x = x_{P} + r cos (13 5^{\circ})

and

y = y_{P} + r sin (13 5^{\circ})

, where

r = \pm \frac{2 2}{3}

. This will give you the coordinates of

Q

and

R

Hint 3

The Execution: For an ellipse

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

, the tangent at

(x_{0}, y_{0})

\frac{x x _{0}}{a ^{2}} + \frac{y y _{0}}{b ^{2}} = 1

. Compare this equation with the given tangent line

x + y = 3

(or

\frac{x}{3} + \frac{y}{3} = 1

) to solve for

a^{2}

and

b^{2}

for both ellipses. Then, calculate the eccentricities

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

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

depending on the major axis.

Visualized Solution

Visualizing the Setup

We are given two ellipses $E_{1}$ and $E_{2}$ centered at the origin.
Ellipse $E_{1}$ has its major axis along the $x$ -axis: $\frac{x ^{2}}{a _{1}^{2}} + \frac{y ^{2}}{b _{1}^{2}} = 1$ with $a_{1} > b_{1}$ .
Ellipse $E_{2}$ has its major axis along the $y$ -axis: $\frac{x ^{2}}{b _{2}^{2}} + \frac{y ^{2}}{a _{2}^{2}} = 1$ with $a_{2} > b_{2}$ .
The circle $S$ is given by $x^{2} + (y - 1)^{2} = 2$ , with center $(0, 1)$ and radius $2$ .
The line $x + y = 3$ is a common tangent to all three curves.

Finding Point of Contact $P$

The tangent line is $x + y = 3$ , which has a slope of $m = - 1$ .
The normal to the circle at the point of contact $P$ must pass through the circle's center $(0, 1)$ .
Since the normal is perpendicular to the tangent, its slope is $m^{'} = 1$ .
Equation of the normal: $y - 1 = 1 (x - 0) \Rightarrow y = x + 1$ .

Solving for Point $P$

We find the intersection of the tangent line $x + y = 3$ and the normal line $y = x + 1$ .
Substitute $y = x + 1$ into the tangent equation:
$x + (x + 1) = 3 \Rightarrow 2 x = 2 \Rightarrow x = 1$ .
Substitute $x = 1$ back to get $y = 2$ .
Thus, the point of contact is $P (1, 2)$ .

Parametric Setup for $Q$ and $R$

The points $Q$ and $R$ lie on the tangent line $x + y = 3$ .
We are given that $P Q = P R = \frac{2 2}{3}$ .
The line $x + y = 3$ has an inclination angle of $θ = 13 5^{\circ}$ (or $\frac{3 π}{4}$ ).
Using parametric form from $P (1, 2)$ :
$x = 1 + r cos (13 5^{\circ})$ and $y = 2 + r sin (13 5^{\circ})$ , where $r = \pm \frac{2 2}{3}$ .

Computing Coordinates of $Q$ and $R$

Substitute $cos (13 5^{\circ}) = - \frac{1}{2}$ and $sin (13 5^{\circ}) = \frac{1}{2}$ :
$x = 1 \pm \frac{2 2}{3} (- \frac{1}{2}) = 1 \mp \frac{2}{3}$
$y = 2 \pm \frac{2 2}{3} (\frac{1}{2}) = 2 \pm \frac{2}{3}$
For $Q$ (further along positive x): $Q\left(\frac{5}{3}, \frac{4}{3} ight)$.
For $R$ (further along positive y): $R\left(\frac{1}{3}, \frac{8}{3} ight)$.

Tangent Condition for Ellipse $E_{1}$

The equation of ellipse $E_{1}$ is $\frac{x ^{2}}{a _{1}^{2}} + \frac{y ^{2}}{b _{1}^{2}} = 1$ .
The tangent to $E_{1}$ at $Q\left(\frac{5}{3}, \frac{4}{3} ight)$ is given by:
$\frac{x ( \frac{5}{3} )}{a _{1}^{2}} + \frac{y ( \frac{4}{3} )}{b _{1}^{2}} = 1 \Rightarrow \frac{5 x}{3 a _{1}^{2}} + \frac{4 y}{3 b _{1}^{2}} = 1$ .
Compare this with the given tangent line: $\frac{x}{3} + \frac{y}{3} = 1$ .

Solving for $a_{1}^{2}$ and $b_{1}^{2}$

Comparing coefficients:
$\frac{5}{3 a _{1}^{2}} = \frac{1}{3} \Rightarrow a_{1}^{2} = 5$ .
$\frac{4}{3 b _{1}^{2}} = \frac{1}{3} \Rightarrow b_{1}^{2} = 4$ .
Note that $a_{1}^{2} > b_{1}^{2}$ , which is consistent with the major axis being along the $x$ -axis.

Tangent Condition for Ellipse $E_{2}$

The equation of ellipse $E_{2}$ is $\frac{x ^{2}}{b _{2}^{2}} + \frac{y ^{2}}{a _{2}^{2}} = 1$ .
The tangent to $E_{2}$ at $R\left(\frac{1}{3}, \frac{8}{3} ight)$ is given by:
$\frac{x ( \frac{1}{3} )}{b _{2}^{2}} + \frac{y ( \frac{8}{3} )}{a _{2}^{2}} = 1 \Rightarrow \frac{x}{3 b _{2}^{2}} + \frac{8 y}{3 a _{2}^{2}} = 1$ .
Compare this with the tangent line: $\frac{x}{3} + \frac{y}{3} = 1$ .

Solving for $a_{2}^{2}$ and $b_{2}^{2}$

Comparing coefficients:
$\frac{1}{3 b _{2}^{2}} = \frac{1}{3} \Rightarrow b_{2}^{2} = 1$ .
$\frac{8}{3 a _{2}^{2}} = \frac{1}{3} \Rightarrow a_{2}^{2} = 8$ .
Note that $a_{2}^{2} > b_{2}^{2}$ , which is consistent with the major axis being along the $y$ -axis.

Calculating Eccentricity $e_{1}$

For ellipse $E_{1}$ , the eccentricity $e_{1}$ is given by:
$e_{1}^{2} = 1 - \frac{b _{1}^{2}}{a _{1}^{2}}$
Substitute $a_{1}^{2} = 5$ and $b_{1}^{2} = 4$ :
$e_{1}^{2} = 1 - \frac{4}{5} = \frac{1}{5}$ .

Calculating Eccentricity $e_{2}$

For ellipse $E_{2}$ , the eccentricity $e_{2}$ is given by:
$e_{2}^{2} = 1 - \frac{b _{2}^{2}}{a _{2}^{2}}$
Substitute $a_{2}^{2} = 8$ and $b_{2}^{2} = 1$ :
$e_{2}^{2} = 1 - \frac{1}{8} = \frac{7}{8}$ .

Verifying Option A

Let's check Option A: $e_{1}^{2} + e_{2}^{2} = \frac{43}{40}$ .
Substitute the values:
$e_{1}^{2} + e_{2}^{2} = \frac{1}{5} + \frac{7}{8} = \frac{8 + 35}{40} = \frac{43}{40}$ .
This matches Option A perfectly! Thus, Option A is correct.

Verifying Option B

Let's check Option B: $e_{1} e_{2} = \frac{7}{2 10}$ .
Calculate the product:
$e_{1} e_{2} = e_{1}^{2} e_{2}^{2} = \frac{1}{5} \cdot \frac{7}{8} = \frac{7}{40}$ .
Simplify the denominator: $40 = 4 \times 10 = 210$ .
So, $e_{1} e_{2} = \frac{7}{2 10}$ .
This matches Option B perfectly! Thus, Option B is also correct.

Final Summary

We successfully determined the points of contact $P$ , $Q$ , and $R$ .
Using the tangent conditions, we found the semi-axes of both ellipses.
The eccentricities were calculated as $e_{1}^{2} = \frac{1}{5}$ and $e_{2}^{2} = \frac{7}{8}$ .
Both Option A and Option B are correct.

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The circle

S

is centered at

(0, 1)

with a radius of

2

. The line

x + y = 3

acts as a common tangent to the circle

S

and two ellipses,

E_{1}

and

E_{2}

To find the point of contact

P

between the line and the circle, we note that the tangent has a slope of

- 1

. The normal to the circle at

P

must have a slope of

1

and pass through the center

(0, 1)

The equation of the normal is

y - 1 = 1 (x - 0)

, which simplifies to

y = x + 1

. Solving the system:

x + (x + 1) = 3 \Rightarrow 2 x = 2 \Rightarrow x = 1

Substituting

x = 1

into the line equation, we find

y = 2

. Thus, the point of contact is

P (1, 2)

Locating Points Q and R

We are given

P Q = P R = \frac{2 2}{3}

. The line

x + y = 3

makes an angle of

13 5^{\circ}

with the positive

x

-axis. Using the parametric form of the line:

(x, y) = (1 + r cos (13 5^{\circ}), 2 + r sin (13 5^{\circ}))

Substituting

r = \pm \frac{2 2}{3}

cos (13 5^{\circ}) = - \frac{1}{2}

, and

sin (13 5^{\circ}) = \frac{1}{2}

x = 1 \mp \frac{2}{3}, y = 2 \pm \frac{2}{3}

This yields the points

Q (\frac{5}{3}, \frac{4}{3})

and

R (\frac{1}{3}, \frac{8}{3})

Determining Ellipse Parameters

For

E_{1}

(major axis on the

x

-axis), the tangent at

(x_{0}, y_{0})

\frac{x x _{0}}{a _{1}^{2}} + \frac{y y _{0}}{b _{1}^{2}} = 1

. Comparing this to the line

\frac{x}{3} + \frac{y}{3} = 1

Q (\frac{5}{3}, \frac{4}{3})

\frac{5}{3 a _{1}^{2}} = \frac{1}{3} \Rightarrow a_{1}^{2} = 5, \frac{4}{3 b _{1}^{2}} = \frac{1}{3} \Rightarrow b_{1}^{2} = 4

For

E_{2}

(major axis on the

y

-axis), the tangent at

(x_{0}, y_{0})

\frac{x x _{0}}{b _{2}^{2}} + \frac{y y _{0}}{a _{2}^{2}} = 1

. Comparing this to the line

\frac{x}{3} + \frac{y}{3} = 1

R (\frac{1}{3}, \frac{8}{3})

\frac{1}{3 b _{2}^{2}} = \frac{1}{3} \Rightarrow b_{2}^{2} = 1, \frac{8}{3 a _{2}^{2}} = \frac{1}{3} \Rightarrow a_{2}^{2} = 8

Final Calculation

The eccentricities are calculated as follows:

e_{1}^{2} = 1 - \frac{b _{1}^{2}}{a _{1}^{2}} = 1 - \frac{4}{5} = \frac{1}{5}

e_{2}^{2} = 1 - \frac{b _{2}^{2}}{a _{2}^{2}} = 1 - \frac{1}{8} = \frac{7}{8}

The sum of the squares of the eccentricities is:

e_{1}^{2} + e_{2}^{2} = \frac{1}{5} + \frac{7}{8} = \frac{8 + 35}{40} = \frac{43}{40}

Common Traps & FAQs

Why can't we just use the condition

c^{2} = a^{2} m^{2} + b^{2}

for the tangent? It seems much faster.

That is a brilliant observation, but it is incomplete for this specific problem. The condition

c^{2} = a^{2} m^{2} + b^{2}

tells us that a line is tangent, but it does not tell us *where* it touches. Since we need to find the specific points

Q

and

R

to define the ellipses, we must use the point of contact method. The condition

c^{2} = a^{2} m^{2} + b^{2}

would only give us a relationship between

a

and

b

, not their individual values.

How do I know which point is

Q

and which is

R

Great question! The problem states

E_{1}

has its major axis along the

x

-axis and

E_{2}

along the

y

-axis. When you calculate the two points using the parametric distance, you will get two sets of coordinates. One set will yield

a_{1}^{2} > b_{1}^{2}

(consistent with

E_{1}

), and the other will yield

a_{2}^{2} > b_{2}^{2}

(consistent with

E_{2}

). The geometry naturally sorts itself out if you follow the constraints of the major axes.

Why is the slope of the normal

1

Remember that the tangent line is

x + y = 3

, which can be rewritten as

y = - x + 3

. The slope of this line is

- 1

. A normal is, by definition, perpendicular to the tangent at the point of contact. If two lines are perpendicular, the product of their slopes is

- 1

. So, if the tangent slope is

- 1

, the normal slope must be

1

because

(- 1) \times (1) = - 1

Select Answer:

Visualized Solution

Visualizing the Setup

Finding Point of Contact $P$

Solving for Point $P$

Parametric Setup for $Q$ and $R$

Computing Coordinates of $Q$ and $R$

Tangent Condition for Ellipse $E_{1}$

Solving for $a_{1}^{2}$ and $b_{1}^{2}$

Tangent Condition for Ellipse $E_{2}$

Solving for $a_{2}^{2}$ and $b_{2}^{2}$

Calculating Eccentricity $e_{1}$

Calculating Eccentricity $e_{2}$

Verifying Option A

Verifying Option B

Final Summary

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Locating Points Q and R

Determining Ellipse Parameters

Final Calculation

Similar Questions

Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{4} = 1$ meet at the point $R (2, 22 - 2)$ . If $S$ is the focus of the ellipse on its negative major axis, then $S P^{2} + S Q^{2}$ is equal to

If the maximum distance of normal to the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{b ^{2}} = 1, b < 2$ , from the origin is 1, then the eccentricity of the ellipse is:

Let the line $y = m x$ and the ellipse $2 x^{2} + y^{2} = 1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(\frac{1}{3 2}, 0)$ and $(0, β)$ , then $β$ is equal to :

If the line $α x + 4 y = 7$ , where $α \in R$ , touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

Finding Point of Contact .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

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

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

Tangent Condition for Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a12​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b12​

Tangent Condition for Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a22​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b22​

Calculating Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e1​

Calculating Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2​

Verifying Option A

Verifying Option B

Final Summary

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Locating Points Q and R

Determining Ellipse Parameters

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the maximum distance of normal to the ellipse 4x2​+b2y2​=1,b<2, from the origin is 1, then the eccentricity of the ellipse is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line αx+4y=7​, where α∈R, touches the ellipse 3x2+4y2=1 at the point P in the first quadrant, then one of the focal distances of P is :

Finding Point of Contact $P$

Solving for Point $P$

Parametric Setup for $Q$ and $R$

Computing Coordinates of $Q$ and $R$

Tangent Condition for Ellipse $E_{1}$

Solving for $a_{1}^{2}$ and $b_{1}^{2}$

Tangent Condition for Ellipse $E_{2}$

Solving for $a_{2}^{2}$ and $b_{2}^{2}$

Calculating Eccentricity $e_{1}$

Calculating Eccentricity $e_{2}$

If the maximum distance of normal to the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{b ^{2}} = 1, b < 2$ , from the origin is 1, then the eccentricity of the ellipse is:

If the line $α x + 4 y = 7$ , where $α \in R$ , touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is :