JEE Main 2019 (8 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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 :

Q: I found $b = -8a$, but why do I need to plug it back into the ellipse equation? Isn't that just a relationship?

Great question! $b = -8a$ is a relationship between the coordinates of the second point. But that point is not just any point in the plane; it is a point *on the ellipse*. To find the specific value of $a^2$, you must satisfy the constraint of the curve itself. It is the intersection of the 'perpendicularity condition' and the 'ellipse constraint'.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry of the ellipse

4 x^{2} + y^{2} = 8

. You are dealing with two tangents. The first is fixed at

(1, 2)

, and the second is variable at

(a, b)

. The core geometric reality is that these two lines are perpendicular, which imposes a strict constraint on their slopes.

Hint 2

Hint 2: Utilize the powerful

T = 0

method to find the equation of the tangent at any point

(x_{1}, y_{1})

on the ellipse, which is

4 x x_{1} + y y_{1} = 8

. Recall that for two lines with slopes

m_{1}

and

m_{2}

to be perpendicular, their product must satisfy

m_{1} \cdot m_{2} = - 1

Hint 3

Hint 3: Calculate the slope of the first tangent at

(1, 2)

to get

m_{1} = - 2

. Then, express the slope of the second tangent at

(a, b)

m_{2} = - \frac{4 a}{b}

. Set their product to

- 1

to derive the relationship

b = - 8 a

. Finally, substitute this relationship into the ellipse equation

4 a^{2} + b^{2} = 8

to solve for

a^{2}

Visualized Solution

Visualizing the Ellipse

Given Ellipse: $4 x^{2} + y^{2} = 8$
Standard Form: $\frac{x ^{2}}{2} + \frac{y ^{2}}{8} = 1$

Tangent at Point $(1, 2)$

Point $P_{1}$ : $(1, 2)$
Tangent equation using $T = 0$ :
$4 x x_{1} + y y_{1} = 8$

Slope of First Tangent

Substitute $(1, 2)$ into $T = 0$ :
$4 (1) x + (2) y = 8 \Rightarrow 2 x + y = 4$
Slope $m_{1} = - 2$

Tangent at Point $(a, b)$

Point $P_{2}$ : $(a, b)$
Tangent equation:
$4 a x + b y = 8$

Slope of Second Tangent

From $4 a x + b y = 8$ :
Slope $m_{2} = - \frac{4 a}{b}$

Applying Perpendicularity

Tangents are perpendicular $\Rightarrow m_{1} \cdot m_{2} = - 1$
Substitute:
$(- 2) \cdot (- \frac{4 a}{b}) = - 1$

Finding Relation between $a$ and $b$

$\frac{8 a}{b} = - 1$
$b = - 8 a$

Point $(a, b)$ on the Ellipse

$(a, b)$ lies on $4 x^{2} + y^{2} = 8$
Substitute $x = a, y = b$ :
$4 a^{2} + b^{2} = 8$

Solving for $a^{2}$

Substitute $b = - 8 a$ into $4 a^{2} + b^{2} = 8$ :
$4 a^{2} + (- 8 a)^{2} = 8$
$4 a^{2} + 64 a^{2} = 8$

Final Conclusion

$68 a^{2} = 8 \Rightarrow a^{2} = \frac{8}{68}$
$a^{2} = \frac{2}{17}$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The ellipse is defined by the equation

4 x^{2} + y^{2} = 8

. To normalize this into standard form, we divide the entire equation by

8

\frac{x ^{2}}{2} + \frac{y ^{2}}{8} = 1

This represents an ellipse stretched along the

y

-axis, serving as our primary coordinate geometry playing field.

The First Tangent

The Known Path

We are given a point

(1, 2)

on the ellipse. To find the tangent line at this point, we utilize the

T = 0

method, where the tangent to

A x^{2} + B y^{2} = C

(x_{1}, y_{1})

is given by

A x x_{1} + B y y_{1} = C

Applying this to

4 x^{2} + y^{2} = 8

(1, 2)

, we obtain:

4 (1) x + (2) y = 8

Simplifying this, we get

2 x + y = 4

, or

y = - 2 x + 4

. The slope of this first tangent,

m_{1}

, is $- 2$ .

The Mystery Point

The Second Tangent

Consider a second point

(a, b)

on the ellipse. Using the same

T = 0

logic, the equation of the tangent at this point is

4 a x + b y = 8

To find the slope

m_{2}

, we rearrange the equation into slope-intercept form:

b y = - 4 a x + 8 \Rightarrow y = - \frac{4 a}{b} x + \frac{8}{b}

Thus, the slope of the second tangent is

m_{2} = - \frac{4 a}{b}

The Bridge

Perpendicularity

The problem states that the two tangents are perpendicular. For any two perpendicular lines, the product of their slopes must satisfy

m_{1} \cdot m_{2} = - 1

Substituting our known values:

(- 2) \cdot (- \frac{4 a}{b}) = - 1

This simplifies to

\frac{8 a}{b} = - 1

, which yields the relationship $b = - 8 a$ .

The Final Synthesis

Since the point

(a, b)

lies on the ellipse

4 x^{2} + y^{2} = 8

, it must satisfy the equation

4 a^{2} + b^{2} = 8

. We substitute our relationship

b = - 8 a

into this equation:

4 a^{2} + (- 8 a)^{2} = 8

4 a^{2} + 64 a^{2} = 8

68 a^{2} = 8

Solving for

a^{2}

, we find:

a^{2} = \frac{8}{68} = \frac{2}{17}

The final value is $a^{2} = \frac{2}{17}$ .

Common Traps & FAQs

Wait, why do we use the

T = 0

method? Can't we just differentiate the ellipse equation to find the slope?

You absolutely can! Differentiation is a fundamental tool. However, the

T = 0

method is a specialized, elegant shortcut for conic sections that saves precious time in the exam hall. It is less prone to algebraic errors and keeps your workspace clean. Both methods lead to the same result, but

T = 0

is the 'pro' way to handle conics.

Why are we setting the product of slopes to

- 1

? Is that always true for tangents?

That condition is specifically for perpendicular lines. The problem explicitly states the tangents are perpendicular. If they were parallel, we would set

m_{1} = m_{2}

. Always read the problem carefully—the geometric condition dictates the algebraic constraint you must apply.

I found

b = - 8 a

, but why do I need to plug it back into the ellipse equation? Isn't that just a relationship?

Great question!

b = - 8 a

is a relationship between the coordinates of the second point. But that point is not just any point in the plane; it is a point *on the ellipse*. To find the specific value of

a^{2}

, you must satisfy the constraint of the curve itself. It is the intersection of the 'perpendicularity condition' and the 'ellipse constraint'.

Animated Solution for Mathematics - Conic Sections: 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 :

Select Answer:

Visualized Solution

Visualizing the Ellipse

Tangent at Point $(1, 2)$

Slope of First Tangent

Tangent at Point $(a, b)$

Slope of Second Tangent

Applying Perpendicularity

Finding Relation between $a$ and $b$

Point $(a, b)$ on the Ellipse

Solving for $a^{2}$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The First Tangent

The Mystery Point

The Bridge

The Final Synthesis

Similar Questions

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

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 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 ____.

A tangent line $L$ is drawn at the point $(2, - 4)$ on the parabola $y^{2} = 8 x$ . If the line $L$ is also tangent to the circle $x^{2} + y^{2} = a$ , then 'a' is equal to ,

Let $L$ be a tangent line to the parabola $y^{2} = 4 x - 20$ at $(6, 2)$ . If $L$ is also a tangent to the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{b} = 1$ , then the value of $b$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Ellipse

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

Slope of First Tangent

Tangent at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a,b)

Slope of Second Tangent

Applying Perpendicularity

Finding Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a,b) on the Ellipse

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The First Tangent

The Mystery Point

The Bridge

The Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the parabola y2=x at a point (α,β),(β>0) is also a tangent to the ellipse, x2+2y2=1, then α is equal to :

.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; } A tangent line L is drawn at the point (2,−4) on the parabola y2=8x. If the line L is also tangent to the circle x2+y2=a, then 'a' is equal to ,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let L be a tangent line to the parabola y2=4x−20 at (6,2). If L is also a tangent to the ellipse 2x2​+by2​=1, then the value of b is equal to :

Tangent at Point $(1, 2)$

Tangent at Point $(a, b)$

Finding Relation between $a$ and $b$

Point $(a, b)$ on the Ellipse

Solving for $a^{2}$

If the tangent to the parabola $y^{2} = x$ at a point $(α, β), (β > 0)$ is also a tangent to the ellipse, $x^{2} + 2 y^{2} = 1$ , then $α$ is equal to :

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:

A tangent line $L$ is drawn at the point $(2, - 4)$ on the parabola $y^{2} = 8 x$ . If the line $L$ is also tangent to the circle $x^{2} + y^{2} = a$ , then 'a' is equal to ,

Let $L$ be a tangent line to the parabola $y^{2} = 4 x - 20$ at $(6, 2)$ . If $L$ is also a tangent to the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{b} = 1$ , then the value of $b$ is equal to :