JEE Main 2020 - 2 Sep (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: A line parallel to the straight line $2 x - y = 0$ is tangent to the hyperbola $\frac{x ^{2}}{4} - \frac{y ^{2}}{2} = 1$ at the point $(x_{1}, y_{1})$ . Then $x_{1}^{2} + 5 y_{1}^{2}$ is equal to :

Q: Why is the slope of the tangent $\frac{1}{\sqrt{2} \sin \theta}$?

Great catch! When you write the tangent equation as $\frac{x \sec \theta}{2} - \frac{y \tan \theta}{\sqrt{2}} = 1$, rearranging it into $y = mx + c$ gives you the slope. By simplifying the ratio of the coefficients of $x$ and $y$, you arrive at that expression. It's just a matter of careful algebraic manipulation!

Q: Is it possible to solve this using the condition of tangency $c^2 = a^2 m^2 - b^2$?

Yes, that is a fantastic alternative! Since the line is $y = 2x + c$, we know $m=2$. You can find $c$ using the condition, then find the point of contact $(x_1, y_1)$ using the formula $(-\frac{a^2 m}{c}, -\frac{b^2}{c})$. Both methods lead to the same beautiful result.

Q: I got a different value for the final expression. Did I mess up the arithmetic?

Don't worry! Check your substitution of $\sec^2 \theta$ and $\tan^2 \theta$. Remember that $\sec^2 \theta = 1 + \tan^2 \theta$. If you accidentally swapped these or made a sign error in the tangent equation, the final sum will deviate. Re-verify the simplification of $4 \sec^2 \theta + 10 \tan^2 \theta$.

Select Answer:

Visualized Solution

$A na l y z e t h eH y p er b o l a$

Given Hyperbola: $\frac{x ^{2}}{4} - \frac{y ^{2}}{2} = 1$
Standard form: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Comparing, we get $a^{2} = 4 \Rightarrow a = 2$ and $b^{2} = 2 \Rightarrow b = 2$

$P a r am e t r i c C oor d ina t es (x_{1}, y_{1})$

Let the point $(x_{1}, y_{1})$ on the hyperbola be in parametric form.
$(x_{1}, y_{1}) = (a sec θ, b tan θ)$
Substituting $a = 2$ and $b = 2$ :
$(x_{1}, y_{1}) = (2 sec θ, 2 tan θ)$

$E q u a t i o n o f t h e T an g e n t$

Equation of tangent at $(x_{1}, y_{1})$ is $\frac{x x _{1}}{a ^{2}} - \frac{y y _{1}}{b ^{2}} = 1$
Substitute $x_{1} = 2 sec θ$ and $y_{1} = 2 tan θ$ :
$\frac{x ( 2 s e c θ )}{4} - \frac{y ( 2 t a n θ )}{2} = 1$
Simplified Tangent Equation: $\frac{x s e c θ}{2} - \frac{y t a n θ}{2} = 1$

$F in d in g t h e S l o p eo f T an g e n t$

Rearrange tangent equation to $y = m x + c$ form:
$\frac{y t a n θ}{2} = \frac{x s e c θ}{2} - 1$
$y = (\frac{2 s e c θ}{2 t a n θ}) x - \frac{2}{t a n θ}$
Slope $m_{t} = \frac{2 \cdot \frac{1}{c o s θ}}{2 \cdot \frac{s i n θ}{c o s θ}} = \frac{1}{2 s i n θ}$

$P a r a l l e l L in e C o n d i t i o n$

Given line: $2 x - y = 0 \Rightarrow y = 2 x$
Slope of given line $m_{l} = 2$
Since tangent is parallel to this line, $m_{t} = m_{l}$
$\frac{1}{2 s i n θ} = 2$

$S o l v in g f or sin θ$

From $\frac{1}{2 s i n θ} = 2$ , we get:
$sin θ = \frac{1}{2 2}$
Squaring both sides:
$sin^{2} θ = \frac{1}{( 2 2 ) ^{2}} = \frac{1}{8}$

$F in d in g sec^{2} θ an d tan^{2} θ$

$cos^{2} θ = 1 - sin^{2} θ = 1 - \frac{1}{8} = \frac{7}{8}$
$sec^{2} θ = \frac{1}{c o s ^{2} θ} = \frac{8}{7}$
$tan^{2} θ = sec^{2} θ - 1 = \frac{8}{7} - 1 = \frac{1}{7}$

$F ina l E x p r ess i o n C a l c u l a t i o n$

Target Expression: $x_{1}^{2} + 5 y_{1}^{2}$
Substitute $x_{1} = 2 sec θ$ and $y_{1} = 2 tan θ$ :
$x_{1}^{2} + 5 y_{1}^{2} = (2 sec θ)^{2} + 5 (2 tan θ)^{2}$
$= 4 sec^{2} θ + 5 (2 tan^{2} θ) = 4 sec^{2} θ + 10 tan^{2} θ$

$T h e F ina l R es u l t$

Substitute $sec^{2} θ = \frac{8}{7}$ and $tan^{2} θ = \frac{1}{7}$ :
$= 4 (\frac{8}{7}) + 10 (\frac{1}{7})$
$= \frac{32}{7} + \frac{10}{7} = \frac{42}{7}$
Final Answer: 6

$K ey T ak e a w a y s$

Key Takeaway: Parametric coordinates $(a sec θ, b tan θ)$ simplify tangent problems.
Key Takeaway: Parallel lines have equal slopes ( $m_{1} = m_{2}$ ).
Next Challenge: What if the line was perpendicular to $2 x - y = 0$ ? How would the value of $x_{1}^{2} + 5 y_{1}^{2}$ change?

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are working with the hyperbola defined by the equation:

\frac{x ^{2}}{4} - \frac{y ^{2}}{2} = 1

We seek a point

(x_{1}, y_{1})

on this curve such that the tangent line at that point is parallel to the line

2 x - y = 0

The Parametric Key

For a hyperbola of the form

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

, we utilize the parametric coordinates

(a sec θ, b tan θ)

. By comparing our equation to the standard form, we identify

a^{2} = 4

(so

a = 2

) and

b^{2} = 2

(so

b = 2

Thus, any point on our hyperbola can be expressed as

(2 sec θ, 2 tan θ)

. This parameter

θ

uniquely defines the position of our point.

The Tangent's Soul

The equation of a tangent at any point

(x_{1}, y_{1})

on the hyperbola is given by:

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

Substituting our parametric coordinates

(x_{1}, y_{1}) = (2 sec θ, 2 tan θ)

, we obtain:

\frac{x ( 2 sec θ )}{4} - \frac{y ( 2 tan θ )}{2} = 1

Simplifying this expression yields:

\frac{x sec θ}{2} - \frac{y tan θ}{2} = 1

Rearranging into the slope-intercept form

y = m x + c

, we find the slope of the tangent

m_{t}

to be:

m_{t} = \frac{1}{2 sin θ}

The Parallel Condition

The given line

2 x - y = 0

has a slope of

2

. Since our tangent must be parallel to this line, we set

m_{t} = 2

\frac{1}{2 sin θ} = 2 ⟹ sin θ = \frac{1}{2 2}

Squaring this result, we obtain

sin^{2} θ = \frac{1}{8}

. We now aim to evaluate the expression

x_{1}^{2} + 5 y_{1}^{2}

Final Calculation

Using our parametric definitions, we have

x_{1}^{2} = 4 sec^{2} θ

and

y_{1}^{2} = 2 tan^{2} θ

. Substituting these into our target expression gives:

4 sec^{2} θ + 10 tan^{2} θ

Given

sin^{2} θ = \frac{1}{8}

, we find

cos^{2} θ = 1 - \frac{1}{8} = \frac{7}{8}

. Consequently,

sec^{2} θ = \frac{8}{7}

and

tan^{2} θ = sec^{2} θ - 1 = \frac{1}{7}

Substituting these values into our expression:

4 (\frac{8}{7}) + 10 (\frac{1}{7}) = \frac{32}{7} + \frac{10}{7} = \frac{42}{7} = 6

The final value is 6.

Common Traps & FAQs

Wait, why do we use

(a sec θ, b tan θ)

instead of just differentiating the equation?

You absolutely can differentiate! However, using parametric coordinates is often the 'elegant' path in JEE problems. It transforms a calculus problem into a trigonometric identity problem, which is usually faster and less prone to algebraic errors during the substitution phase.

Why is the slope of the tangent

\frac{1}{2 s i n θ}

Great catch! When you write the tangent equation as

\frac{x s e c θ}{2} - \frac{y t a n θ}{2} = 1

, rearranging it into

y = m x + c

gives you the slope. By simplifying the ratio of the coefficients of

x

and

y

, you arrive at that expression. It's just a matter of careful algebraic manipulation!

Is it possible to solve this using the condition of tangency

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

Yes, that is a fantastic alternative! Since the line is

y = 2 x + c

, we know

m = 2

. You can find

c

using the condition, then find the point of contact

(x_{1}, y_{1})

using the formula

(- \frac{a ^{2} m}{c}, - \frac{b ^{2}}{c})

. Both methods lead to the same beautiful result.

Why did we get

sin^{2} θ = 1/8

? Does this mean there are multiple points?

Exactly! The hyperbola is symmetric. A line with a given slope can be tangent to either the right branch or the left branch of the hyperbola. The squared value accounts for these symmetric possibilities, ensuring our final expression

x_{1}^{2} + 5 y_{1}^{2}

remains consistent regardless of which branch we are on.

I got a different value for the final expression. Did I mess up the arithmetic?

Don't worry! Check your substitution of

sec^{2} θ

and

tan^{2} θ

. Remember that

sec^{2} θ = 1 + tan^{2} θ

. If you accidentally swapped these or made a sign error in the tangent equation, the final sum will deviate. Re-verify the simplification of

4 sec^{2} θ + 10 tan^{2} θ

Animated Solution for Mathematics - Conic Sections: A line parallel to the straight line $2 x - y = 0$ is tangent to the hyperbola $\frac{x ^{2}}{4} - \frac{y ^{2}}{2} = 1$ at the point $(x_{1}, y_{1})$ . Then $x_{1}^{2} + 5 y_{1}^{2}$ is equal to :

Select Answer:

Visualized Solution

$A na l y z e t h eH y p er b o l a$

$P a r am e t r i c C oor d ina t es (x_{1}, y_{1})$

$E q u a t i o n o f t h e T an g e n t$

$F in d in g t h e S l o p eo f T an g e n t$

$P a r a l l e l L in e C o n d i t i o n$

$S o l v in g f or sin θ$

$F in d in g sec^{2} θ an d tan^{2} θ$

$F ina l E x p r ess i o n C a l c u l a t i o n$

$T h e F ina l R es u l t$

$K ey T ak e a w a y s$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Key

The Tangent's Soul

The Parallel Condition

Final Calculation

Similar Questions

The equation of a tangent to the hyperbola $4 x^{2} - 5 y^{2} = 20$ parallel to the line $x - y = 2$ is :

If a hyperbola passes through the point $P (10, 16)$ and it has vertices at $(\pm 6, 0)$ , then the equation of the normal to it at $P$ is :

Let $C$ be the locus of the mirror image of a point on the parabola $y^{2} = 4 x$ with respect to the line $y = x$ . Then the equation of tangent to $C$ at $P (2, 1)$ is :

Tangents are drawn to the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ , parallel to the straight line $2 x - y = 1$ . The points of contact of the tangents on the hyperbola are

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 the line $2 x + 6 y = 2$ touches the hyperbola $x^{2} - 2 y^{2} = 4$ , then the point of contact is

If the line $y = m x + c$ is a common tangent to the hyperbola $\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1$ and the circle $x^{2} + y^{2} = 36$ , then which one of the following is true?

On the ellipse $4 x^{2} + 9 y^{2} = 1$ , the points at which the tangents are parallel to the line $8 x = 9 y$ are

The equation of a tangent to the parabola $y^{2} = 8 x$ is $y = x + 2$ . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ParametricCoordinates(x1​,y1​)

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingsec2θandtan2θ

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

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

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

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Parametric Key

The Tangent's Soul

The Parallel Condition

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a tangent to the hyperbola 4x2−5y2=20 parallel to the line x−y=2 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a hyperbola passes through the point P(10,16) and it has vertices at (±6,0), then the equation of the normal to it at P is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let C be the locus of the mirror image of a point on the parabola y2=4x with respect to the line y=x. Then the equation of tangent to C at P(2,1) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangents are drawn to the hyperbola 9x2​−4y2​=1, parallel to the straight line 2x−y=1. The points of contact of the tangents on the hyperbola are

.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 the line 2x+6​y=2 touches the hyperbola x2−2y2=4, then the point of contact is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line y=mx+c is a common tangent to the hyperbola 100x2​−64y2​=1 and the circle x2+y2=36, then which one of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } On the ellipse 4x2+9y2=1, the points at which the tangents are parallel to the line 8x=9y are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a tangent to the parabola y2=8x is y=x+2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

$A na l y z e t h eH y p er b o l a$

$P a r am e t r i c C oor d ina t es (x_{1}, y_{1})$

$E q u a t i o n o f t h e T an g e n t$

$F in d in g t h e S l o p eo f T an g e n t$

$P a r a l l e l L in e C o n d i t i o n$

$S o l v in g f or sin θ$

$F in d in g sec^{2} θ an d tan^{2} θ$

$F ina l E x p r ess i o n C a l c u l a t i o n$

$T h e F ina l R es u l t$

$K ey T ak e a w a y s$

The equation of a tangent to the hyperbola $4 x^{2} - 5 y^{2} = 20$ parallel to the line $x - y = 2$ is :

If a hyperbola passes through the point $P (10, 16)$ and it has vertices at $(\pm 6, 0)$ , then the equation of the normal to it at $P$ is :

Let $C$ be the locus of the mirror image of a point on the parabola $y^{2} = 4 x$ with respect to the line $y = x$ . Then the equation of tangent to $C$ at $P (2, 1)$ is :

Tangents are drawn to the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ , parallel to the straight line $2 x - y = 1$ . The points of contact of the tangents on the hyperbola are

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 the line $2 x + 6 y = 2$ touches the hyperbola $x^{2} - 2 y^{2} = 4$ , then the point of contact is

If the line $y = m x + c$ is a common tangent to the hyperbola $\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1$ and the circle $x^{2} + y^{2} = 36$ , then which one of the following is true?

On the ellipse $4 x^{2} + 9 y^{2} = 1$ , the points at which the tangents are parallel to the line $8 x = 9 y$ are

The equation of a tangent to the parabola $y^{2} = 8 x$ is $y = x + 2$ . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is