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JEE Mains Maths · Conic Sections

Tangents and Normals to a Hyperbola

When a line touches, cuts or misses a hyperbola, the tangent in point, parametric and slope form, tangents from a point, the chord with a given midpoint, and the normal.

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Why this matters

Fifteen PYQs. Two formulas carry them: the tangency condition c² = a²m² − b², and the normal a²x/x₁ + b²y/y₁ = a² + b². Two ideas cover the page.

Concept 1 of 2: Tangents, and when a line meets a hyperbola

Everything mirrors the ellipse with b2b^2 changed to −b2-b^2. A line y=mx+cy=mx+c touches x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 when c2=a2m2−b2c^2=a^2m^2-b^2. If c2c^2 is smaller than that, and the slope is steeper than the asymptotes, the line misses the curve altogether. The chord with a given midpoint is again T=S1T=S_1.

Definition

  • Tangency: c2=a2m2−b2c^2=a^2m^2-b^2; the slope form is y=mx±a2m2−b2y=mx\pm\sqrt{a^2m^2-b^2}.
  • At (x1,y1)(x_1,y_1): xx1a2−yy1b2=1\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1. At θ\theta: xsec⁡θa−ytan⁡θb=1\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1.
  • Misses the curve: ∣m∣>ba|m|>\frac{b}{a} and c2<a2m2−b2c^2<a^2m^2-b^2.
  • Tangents from (h,k)(h,k): slopes solve (h2−a2)m2−2hkm+(k2+b2)=0(h^2-a^2)m^2-2hkm+(k^2+b^2)=0.
  • Chord with midpoint (h,k)(h,k): T=S1T=S_1; slope b2ha2k\frac{b^2h}{a^2k}.
  • Feet of perpendiculars from the centre to tangents lie on (x2+y2)2=a2x2−b2y2(x^2+y^2)^2=a^2x^2-b^2y^2.

Tangency condition

y=mx+c touches x2a2−y2b2=1  ⟺  c2=a2m2−b2y=mx+c\ \text{touches}\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\iff c^2=a^2m^2-b^2

Worked example

Find the tangents of slope 22 to x29−y216=1\frac{x^2}{9}-\frac{y^2}{16}=1.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 June 2022 · Q73Moderate

Example 1 · Conic Sections · Tangents and Normals to a Hyperbola

Let the eccentricity of the hyperbola H:x2a2−y2 b2=1H:\frac{x^{2}}{a^{2}}-\frac{y^{2}}{{\text{ }b}^{2}}= 1 be 52\sqrt{\frac{5}{2}} and length of its latus rectum be 626\sqrt{2}, If y=2x+cy = 2x + c is a tangent to the hyperbola HH, then the value of c2c^{2} is equal to

Minus b2b^2 for the hyperbola

For the ellipse c2=a2m2+b2c^2=a^2m^2+b^2; for the hyperbola c2=a2m2−b2c^2=a^2m^2-b^2. A slope with a2m2<b2a^2m^2<b^2 gives no tangent at all.
Drill 9 more on tangents, and when a line meets a hyperbola

Concept 2 of 2: The normal to a hyperbola

The normal at (x1,y1)(x_1,y_1) is perpendicular to the tangent there. Its equation has the same shape as the ellipse's normal with the sign of b2b^2 flipped: a2xx1+b2yy1=a2+b2\frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2. Questions ask where it crosses an axis, or which given point it passes through.

Definition

  • At (x1,y1)(x_1,y_1): a2xx1+b2yy1=a2+b2\frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2.
  • At θ\theta, (asec⁡θ, btan⁡θ)(a\sec\theta,\,b\tan\theta): axcos⁡θ+bycot⁡θ=a2+b2ax\cos\theta+by\cot\theta=a^2+b^2.
  • Slope of the normal at (x1,y1)(x_1,y_1): −a2y1b2x1-\frac{a^2y_1}{b^2x_1}.
  • It meets the transverse axis at x=(a2+b2)x1a2=e2x1x=\frac{(a^2+b^2)x_1}{a^2}=e^2x_1.

Normal at (x₁, y₁)

a2xx1+b2yy1=a2+b2\frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2

Worked example

Find the normal to x29−y216=1\frac{x^2}{9}-\frac{y^2}{16}=1 at (5,163)\left(5,\frac{16}{3}\right).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 25 June 2022 · Q179Moderate

Example 2 · Conic Sections · Tangents and Normals to a Hyperbola

Let the eccentricity of the hyperbola x2a2−y2b2=1\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}= 1 be 54\frac{5}{4}. If the equation of the normal at the point (85,125)\left( \frac{8}{\sqrt{5}},\frac{12}{5} \right) on the hyperbola is 85x+βy=λ8\sqrt{5}x+\beta y=\lambda, then λ−β\lambda-\beta is equal to

PLUS in the normal, MINUS in the tangent

The tangent is xx1a2−yy1b2=1\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1 and the normal is a2xx1+b2yy1=a2+b2\frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2. The signs flip between them.
Drill 4 more on the normal to a hyperbola

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Tangents, and when a line meets a hyperbola

    Tangency condition

    y=mx+c touches x2a2−y2b2=1  ⟺  c2=a2m2−b2y=mx+c\ \text{touches}\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\iff c^2=a^2m^2-b^2
  • The normal to a hyperbola

    Normal at (x₁, y₁)

    a2xx1+b2yy1=a2+b2\frac{a^2x}{x_1}+\frac{b^2y}{y_1}=a^2+b^2

Watch out for (2)

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