Intersection of Line and the Hyperbola

A line may meet the Hyperbola in one point or two distinct points or it may not meet the Hyperbola at all. If the line meets the Hyperbola at one point is called Tangent and If the line meets the hyperbola at two points it is called a chord. In real life, we use tangents in the construction and navigation field to calculate distances, heights, and angles.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas  | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics 

This Story also Contains

1. What is the Hyperbola?
2. Equation of Hyperbola
3. Intersection of Hyperbola and Line
4. Solved Examples Based on Line and the Hyperbola
5. Summary

In this article, we will cover the concept of Line and Hyperbola. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of twelve questions have been asked on JEE MAINS( 2013 to 2023) from this topic including two in 2022.

What is the Hyperbola?

The locus of a point moves in a plane such that the ratio of the distance from a fixed point (focus) to the distance from a fixed line (directrix) is constant. The constant is known as eccentricity e and for hyperbola e > 1.

Equation of Hyperbola

The standard form of the equation of a hyperbola with centre (0, 0) and foci lying on the x-axis is
x2a2−y2b2=1 where, b2=a2(e2−1)

Intersection of Hyperbola and Line

Hyperbola: x2a2−y2b2=1
Line: y=mx+c
After solving Eq. (i) and Eq. (ii)

x2a2−(mx+c)2b2=1⇒(a2m2−b2)x2+2ma2x+c2a2+a2b2=0

The above equation is quadratic in x

The line will cut the hyperbola in two points may be real, coincident or imaginary, depending on the value of Discriminant, D.

Case 1: If D > 0, then two real and distinct roots which means two real and distinct points of intersection of the line and the hyperbola. In this case, the line is secant (chord) to the hyperbola.

Case 2: If D = 0, then equal real roots which means the line is tangent to the hyperbola. Solving D = 0 we get the condition for tangency, which is c2=a2m2−b2

Case 3: If D < 0, then no real root which means the line and hyperbola do not intersect.

Recommended Video Based on Line and Hyperbola

Solved Examples Based on Line and the Hyperbola

Example 1: Let the eccentricity of the hyperbola H:x2a2−y2b2=1 be 52 and length of its latus rectum be 62, If y=2x+c is a tangent to the hyperbola H, then the value of c2 is equal to
[JEE MAINS 2022]
Solution

e2=1+b2a2=52⇒b2a2=32−−(1)2 b2a2=62⇒2 b2=62a⇒3a2=62a⇒a=22, b=23

The equation of the tangent of the slope m=2 is

y=2x+a2×22−b2y=2x+32−12c=20⇒c2=20
Hence, the answer is 20

Example 2: Let λx−2y=μ be a tangent to the hyperbola a2x2−y2=b2. Then (λa)2−(μb)2 is equal to:
[JEE MAINS 2022]
Solution

x2(b2a2)−y2b2=1
Tangent: y=λ2x−μ2
Using condition for tangency for hyperbola:

μ24=b2a2×λ24−b2μ24 b2=14(λ2a2−4)λ2a2−μ2 b2=4
Hence, the answer is 4

Example 3: The locus of a point P(α,β) moving under the condition that the line y=αx+β is a tangent to the hyperbola x2a2−y2b2=1 is
Solution: Condition for Tangency in Hyperbola - C2=a2m2−b2
For the Hyperbola x2a2−y2b2=1
y=αx+β is tangent to x2a2−y2b2=1
Condition for tangency for y=mx+C is C2=a2 m2−b2
i.e. β2=a2α2−b2

Thus replace (α,β) by (x,y)
y2=a2x2−b2 which is hyperbola.
Hence, the answer is a hyperbola

Hence, the answer is a hyperbola

Example 4: If P(θ1) and Q(θ2) are the extremities of any focal chord of the hyperbola x2a2−y2b2=1, then cos2⁡θ1+θ22=λcos2⁡θ1−θ22, where λ is equal to
Solution: Equation of any chord joining the points P(θ1) and Q(θ2) is, xacos⁡(θ1−θ22)−yb
sin⁡(θ1+θ22)=cos⁡(θ1+θ22)If it passes through (ae, 0 ), then

⇒e2cos2⁡(θ1−θ22)=cos2⁡(θ1+θ22)⇒λ=e2=1+b2a2=a2+b2a2
Hence, the answer is a2+b2a2

Example 5: If (asec⁡θ,btan⁡θ) and (asecϕ,btan⁡ϕ) be the coordinate of the ends of a focal chord of x2a2−y2 b2=1, then tan⁡θ2tan⁡ϕ2 equals to
Solution: The equation of the chord connecting the points (asecθ,btan⁡θ) and (asecϕ,btan⁡ϕ) is

xacos⁡(θ+ϕ2)−ybsin⁡(θ+ϕ2)=cos⁡(θ−ϕ2)

If it passes through (ae,0); we have, cos⁡(θ−ϕ2)=cos⁡(θ+ϕ2)

e=cos⁡(θ+ϕ2)cos⁡(θ−ϕ2)=1−tan⁡θ2⋅tan⁡ϕ21+tan⁡θ2tan⁡ϕ2⇒tan⁡θ2⋅tan⁡ϕ2=1−e1+e
Hence, the answer is 1−e1+e

Summary

Based on the point of contact of the line on the Hyperbola, we can differentiate it. So, knowledge of the point of contact of the line with respect to the Hyperbola is necessary for understanding its properties and characteristics. Understanding of point of contact of the line on the Hyperbola is necessary for solving theoretical as well as real-life-based problems.