JEE Main 2021 (27 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $E$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3, - 4)$ , one focus at $(4, - 4)$ and one vertex at $(5, - 4)$ . If $m x - y = 4, m > 0$ is a tangent to the ellipse $E$ , then the value of $5 m^{2}$ is equal to

Q: Is there a way to solve this using the discriminant method instead of the tangency condition?

Yes, you could substitute $y = mx - 4$ into the ellipse equation and set the discriminant of the resulting quadratic equation in $x$ to zero. However, this is computationally heavy and prone to calculation errors. The condition of tangency is a derived result of that exact process, so using it is faster, cleaner, and more reliable.

Enter Numerical Value:

Visualized Solution

Plotting the Given Points

Center $C = (3, - 4)$
Focus $S = (4, - 4)$
Vertex $A = (5, - 4)$

Identifying the Major Axis

The $y$ -coordinates of $C$ , $S$ , and $A$ are all $- 4$ .
Therefore, the major axis is horizontal: $y = - 4$ .

Calculating Semi-major Axis $a$

Distance from Center to Vertex: $a = C A$
$a = ∣5 - 3∣ = 2$

Calculating Eccentricity $e$

Distance from Center to Focus: $a e = C S$
$a e = ∣4 - 3∣ = 1$
Eccentricity $e = \frac{a e}{a} = \frac{1}{2}$

Finding the Semi-minor Axis $b^{2}$

Relation: $b^{2} = a^{2} (1 - e^{2})$
Substitute: $b^{2} = 4 (1 - (\frac{1}{2})^{2})$
Calculate: $b^{2} = 4 (1 - \frac{1}{4}) = 3$

Writing the Ellipse Equation

Standard form: $\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1$
Equation: $\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 4 ) ^{2}}{3} = 1$

Introducing the Tangent Line

Given tangent line: $m x - y = 4$
Rearranging: $y = m x - 4$

Shifting the Origin

To simplify, shift the origin to the center of the ellipse $(3, - 4)$ .
Let $X = x - 3$ and $Y = y + 4$ .
This means $x = X + 3$ and $y = Y - 4$ .

Transforming the Tangent Equation

Substitute $x = X + 3$ and $y = Y - 4$ into $y = m x - 4$ .
$(Y - 4) = m (X + 3) - 4$
$Y = m X + 3 m$

Applying the Condition of Tangency

For an ellipse $\frac{X ^{2}}{a ^{2}} + \frac{Y ^{2}}{b ^{2}} = 1$ , a line $Y = m X + c$ is a tangent if:
$c^{2} = a^{2} m^{2} + b^{2}$

Substituting Values into Tangency Condition

From our transformed line, $c = 3 m$ .
We know $a^{2} = 4$ and $b^{2} = 3$ .
Substitute: $(3 m)^{2} = 4 m^{2} + 3$

Solving for $5 m^{2}$

Expand: $9 m^{2} = 4 m^{2} + 3$
Rearrange: $9 m^{2} - 4 m^{2} = 3$
Simplify: $5 m^{2} = 3$

Final Answer

The value of $5 m^{2}$ is $3$ .

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a set of points: a center at

C (3, - 4)

, a focus at

F (4, - 4)

, and a vertex at

A (5, - 4)

Because the

y

-coordinates are identical, we immediately see that these points lie on a perfectly horizontal line,

y = - 4

. This is the major axis of our ellipse, the line along which the ellipse stretches its wings.

Defining the Parameters

To define this ellipse, we need its dimensions. The distance from the center

C (3, - 4)

to the vertex

A (5, - 4)

is the semi-major axis,

a

A simple calculation,

a = ∣5 - 3∣ = 2

, gives us the horizontal reach. Next, we look at the focus. The distance from the center to the focus is

a e

Here,

a e = ∣4 - 3∣ = 1

. Since we know

a = 2

, the eccentricity

e

is clearly

e = \frac{1}{2}

Now, we need the vertical reach, the semi-minor axis

b

. The relationship between these parameters is a cornerstone of conic sections:

b^{2} = a^{2} (1 - e^{2})

Substituting our values, we get:

b^{2} = 4 (1 - (\frac{1}{2})^{2}) = 4 (1 - \frac{1}{4}) = 3

With

a^{2} = 4

and

b^{2} = 3

, the equation of our ellipse is:

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 4 ) ^{2}}{3} = 1

The Art of Transformation

Now, we face the tangent line

m x - y = 4

. To make our lives easier, we perform a coordinate shift.

By defining

X = x - 3

and

Y = y + 4

, we move the center of the ellipse to the origin

(0, 0)

. This is like changing your perspective to see the problem from the center of the action.

Substituting

x = X + 3

and

y = Y - 4

into our line equation

y = m x - 4

, we get:

(Y - 4) = m (X + 3) - 4

This simplifies beautifully to

Y = m X + 3 m

The Final Convergence

We are now ready for the final act. For an ellipse

\frac{X ^{2}}{a ^{2}} + \frac{Y ^{2}}{b ^{2}} = 1

, a line

Y = m X + c

is tangent if and only if

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

In our shifted system,

c = 3 m

a^{2} = 4

, and

b^{2} = 3

. Plugging these into our condition, we get:

(3 m)^{2} = 4 m^{2} + 3

This simplifies to

9 m^{2} = 4 m^{2} + 3

, which leads to

5 m^{2} = 3

And there it is! The algebra collapses into the answer we sought. The value of

m^{2}

is $\frac{3}{5}$ .

Common Traps & FAQs

Why did we shift the origin? Can't we just use the original equation?

You absolutely could use the original equation, but it would involve a much more tedious algebraic expansion. By shifting the origin to the center

(3, - 4)

, we simplify the ellipse equation to

\frac{X ^{2}}{a ^{2}} + \frac{Y ^{2}}{b ^{2}} = 1

. This allows us to use the standard, elegant condition of tangency

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

directly. It is all about choosing the path of least resistance in competitive exams!

How do I know if the major axis is horizontal or vertical?

Look at the coordinates provided. If the y-coordinates of the center, focus, and vertex are identical, the major axis must be horizontal (parallel to the x-axis). If the x-coordinates were identical, it would be vertical. Always sketch the points first; it prevents silly orientation errors.

Is there a way to solve this using the discriminant method instead of the tangency condition?

Yes, you could substitute

y = m x - 4

into the ellipse equation and set the discriminant of the resulting quadratic equation in

x

to zero. However, this is computationally heavy and prone to calculation errors. The condition of tangency is a derived result of that exact process, so using it is faster, cleaner, and more reliable.

Enter Numerical Value:

Visualized Solution

Plotting the Given Points

Identifying the Major Axis

Calculating Semi-major Axis $a$

Calculating Eccentricity $e$

Finding the Semi-minor Axis $b^{2}$

Writing the Ellipse Equation

Introducing the Tangent Line

Shifting the Origin

Transforming the Tangent Equation

Applying the Condition of Tangency

Substituting Values into Tangency Condition

Solving for $5 m^{2}$

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Defining the Parameters

The Art of Transformation

The Final Convergence

Similar Questions

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ for some $a \in R$ , then the distance between the foci of the ellipse is:

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :

If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point P on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

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 two tangents drawn from a point $(α, β)$ lying on the ellipse $25 x^{2} + 4 y^{2} = 1$ to the parabola $y^{2} = 4 x$ are such that the slope of one tangent is four times the other, then the value of $(10 α + 5)^{2} + (16 β^{2} + 50)^{2}$ equals ____.

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :

Enter Numerical Value:

Visualized Solution

Plotting the Given Points

Identifying the Major Axis

Calculating Semi-major Axis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

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

Finding the Semi-minor Axis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Writing the Ellipse Equation

Introducing the Tangent Line

Shifting the Origin

Transforming the Tangent Equation

Applying the Condition of Tangency

Substituting Values into Tangency Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5m2

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Defining the Parameters

The Art of Transformation

The Final Convergence

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 3x+4y=122​ is a tangent to the ellipse a2x2​+9y2​=1 for some a∈R, then the distance between the foci of the ellipse is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 3x+4y=122​ is a tangent to the ellipse a2x2​+9y2​=1, for some a∈R then the distance between the foci of the ellipse is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normal to the ellipse 3x2+4y2=12 at a point P on it is parallel to the line, 2x+y=4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents on the ellipse 4x2+y2=8 at the points (1,2) and (a,b) are perpendicular to each other, then a2 is equal to :

Calculating Semi-major Axis $a$

Calculating Eccentricity $e$

Finding the Semi-minor Axis $b^{2}$

Solving for $5 m^{2}$

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ for some $a \in R$ , then the distance between the foci of the ellipse is:

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :

If the normal to the ellipse $3 x^{2} + 4 y^{2} = 12$ at a point P on it is parallel to the line, $2 x + y = 4$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :

If the tangents on the ellipse $4 x^{2} + y^{2} = 8$ at the points $(1, 2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal to :