MathsMediumClass 12

Tangent & Normal

Conic Sections

JEE Qs

Hard

min

Master the direct formulas for tangents and normals for all conics in point, parametric, and slope forms, as well as the 'T=0' method, to efficiently solve related problems.

🧮 Key Formulas

General tangent equation at (x1, y1) for a 2nd degree curve S=0: T=0 (replace x^2 with x x1, y^2 with y y1, x with (x+x1)/2, y with (y+y1)/2, xy with (xy1+yx1)/2)

Slope of tangent at (x1, y1) = (dy/dx) at (x1, y1)

Slope of normal at (x1, y1) = -1 / (dy/dx) at (x1, y1)

Equation of tangent at (x1, y1): y - y1 = (dy/dx)(x - x1)

Equation of normal at (x1, y1): y - y1 = (-1 / (dy/dx))(x - x1)

Parabola y^2 = 4ax:

Tangent at (x1, y1): y y1 = 2a(x + x1)

Tangent in parametric form (at^2, 2at): yt = x + at^2

Tangent in slope form y = mx + a/m, point of contact (a/m^2, 2a/m)

Normal at (x1, y1): y - y1 = (-y1 / 2a)(x - x1)

Normal in parametric form: y + tx = 2at + at^3

Normal in slope form y = mx - 2am - am^3, point of contact (am^2, -2am)

Ellipse x^2/a^2 + y^2/b^2 = 1:

Tangent at (x1, y1): x x1/a^2 + y y1/b^2 = 1

Tangent in parametric form (a cosθ, b sinθ): x/a cosθ + y/b sinθ = 1

Tangent in slope form y = mx ± sqrt(a^2m^2 + b^2)

Normal at (x1, y1): a^2x/x1 - b^2y/y1 = a^2 - b^2

Normal in parametric form: ax secθ - by cosecθ = a^2 - b^2

Hyperbola x^2/a^2 - y^2/b^2 = 1:

Tangent at (x1, y1): x x1/a^2 - y y1/b^2 = 1

Tangent in parametric form (a secθ, b tanθ): x/a secθ - y/b tanθ = 1

Tangent in slope form y = mx ± sqrt(a^2m^2 - b^2)

Normal at (x1, y1): a^2x/x1 + b^2y/y1 = a^2 + b^2

Normal in parametric form: ax cosθ + by cotθ = a^2 + b^2

Circle x^2 + y^2 = r^2:

Tangent at (x1, y1): x x1 + y y1 = r^2

Tangent in slope form y = mx ± r sqrt(1+m^2)

Condition for y = mx + c to be tangent to conic:

Parabola y^2 = 4ax: c = a/m

Ellipse x^2/a^2 + y^2/b^2 = 1: c^2 = a^2m^2 + b^2

Hyperbola x^2/a^2 - y^2/b^2 = 1: c^2 = a^2m^2 - b^2

✅ Key Points for JEE

1The general method of finding dy/dx from the conic's equation is universally applicable for finding slopes of tangent and normal at any point, even if specific formulas are forgotten.
2The 'T=0' method provides the tangent equation at a point (x1, y1) on any second-degree curve directly, minimizing calculation errors and saving time.
3Parametric forms of tangent and normal equations are often more convenient for problems involving properties like points of intersection or loci, where a parameter simplifies algebraic manipulation.
4Understand the condition for tangency (y = mx + c touching a conic) as it helps find the equation of tangent when only its slope or an external point is known.
5Many problems involve finding tangents/normals from an external point; in such cases, assume the point of contact as (x1, y1) or (at^2, 2at) and use the tangent equation, then force it through the external point.

⚠️ Common Mistakes

✕Sign errors, especially in hyperbola equations and normal equations (e.g., confusing `a^2 - b^2` with `a^2 + b^2`).
✕Interchanging the slope of the tangent with the slope of the normal (slope of normal = -1/slope of tangent).
✕Incorrectly applying the T=0 method for the chord of contact or polar, or not recognizing when T=0 can be used.
✕Forgetting to check the condition for tangency when a general line y = mx + c is given to touch a conic, leading to incomplete solutions.

📝 Practice Questions

See all

Q70.Let S = {(x, 1}, where (1) An ellipse whose eccentricity is 1 , when (2) A hyperbola whose eccentricity is 2 , when √r+1 √r+1 r > 1. 0 < r < 1. (3) (4) A hyperbola whose eccentricity is 2 , when An ellipse whose eccentricity is , when √1−r √ r+12 r > 1 0 < r < 1

2019·MCQHard

Q69.If the common tangents to the parabola, x2 = 4y and the circle, x2 + y2 = 4 intersect at the point P , then the distance of P from the origin (units), is: + (1) 2(3 2√2) (2) 3 + 2√2 + (3) √2 + 1 (4) 2(√2 1) JEE Main 2017 (08 Apr Online) JEE Main Previous Year Paper

2017·Multi conceptHard

NCERT Chapters

Class 11 Maths Ch 11: Conic Sections
Class 12 Maths Ch 6: Application of Derivatives