JEE Mains Maths · Conic Sections

Tangents, Normals and Chords of an Ellipse

Lines and the ellipse: the tangent in point, parametric and slope form, pairs of tangents and the director circle, the normal, and chords fixed by a midpoint or through a given point.

All Conic Sections notes JEE Mains Maths strategy

Why this matters

Twenty-six PYQs. The tangency condition c² = a²m² + b² and the midpoint-chord rule T = S₁ between them answer most. Four ideas cover the page.

Concept 1 of 4: The tangent: point, parametric and slope forms

The tangent at a point of the ellipse follows the same

T=0

pattern as for a circle: replace

x^2

xx_1

and

y^2

yy_1

. A line

y=mx+c

is a tangent exactly when

c^2=a^2m^2+b^2

. The tangent at the parametric point cuts the axes at

a\sec\theta

and

b\csc\theta

, which makes 'smallest triangle with the axes' questions one line.

Definition

At $(x_1,y_1)$ : $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$ .
At $\theta$ : $\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$ .
Slope form: $y=mx\pm\sqrt{a^2m^2+b^2}$ , touching at $\left(-\frac{a^2m}{c},\frac{b^2}{c}\right)$ .
Intercepts of the tangent at $\theta$ : $a\sec\theta$ and $b\csc\theta$ ; the triangle with the axes has area $\frac{ab}{\sin2\theta}\geq ab$ .

Tangency condition

y=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

1

\frac{x^2}{16}+\frac{y^2}{9}=1

and the contact point of the upper one.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 17 · Q84Moderate

Example 1 · Conic Sections · Tangents, Normals and Chords of an Ellipse

Let

E

be an ellipse whose axes are parallel to the co-ordinate's axes, having its center at

(3, - 4)

, one focus at

(4, - 4)

and one vertex at

(5, - 4)

. If

mx - y = 4,\text{ }m > 0

is a tangent to the ellipse

E

, then the value of

5{\text{ }m}^{2}

is equal to

Shift the line before using $c^2=a^2m^2+b^2$

The condition is for an ellipse centred at the origin. For

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

, rewrite the line in

X=x-h

Y=y-k

first; the slope stays, the intercept changes.

Drill 9 more on the tangent: point, parametric and slope forms

Concept 2 of 4: Pairs of tangents, the chord of contact and the director circle

Tangents from an outside point

(h,k)

have slopes that solve a quadratic, found by making

y-k=m(x-h)

satisfy the tangency condition. When the product of those slopes is

-1

, the point sits on the director circle

x^2+y^2=a^2+b^2

: every pair of perpendicular tangents meets there.

Definition

Slopes from $(h,k)$ : $(h^2-a^2)m^2-2hkm+(k^2-b^2)=0$ .
Chord of contact: $\frac{hx}{a^2}+\frac{ky}{b^2}=1$ .
Pair of tangents: $SS_1=T^2$ .
Director circle: $x^2+y^2=a^2+b^2$ , where perpendicular tangents meet.
Angle $\theta$ between the tangents: $\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$ from the quadratic's sum and product.

Director circle

x^2+y^2=a^2+b^2

Worked example

Where do perpendicular tangents to

\frac{x^2}{16}+\frac{y^2}{9}=1

meet?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q161Moderate

Example 2 · Conic Sections · Tangents, Normals and Chords of an Ellipse

The acute angle between the pair of tangents drawn to the ellipse

2x^{2}+ 3y^{2}= 5

from the point

(1,3)

The director circle uses $a^2+b^2$ , not $a^2$

x^2+y^2=a^2

is the auxiliary circle, the circle on the major axis. The meeting points of perpendicular tangents are farther out, at radius

\sqrt{a^2+b^2}

Drill 1 more on pairs of tangents, the chord of contact and the director circle

Concept 3 of 4: The normal to an ellipse

The normal at a point is perpendicular to the tangent there, with slope

\frac{a^2y_1}{b^2x_1}

. In parametric form it is

ax\sec\theta-by\csc\theta=a^2-b^2

. Two uses come up: where a normal crosses an axis, and the largest circle centred on the major axis that fits inside the ellipse, which touches it where a normal passes through the centre of that circle.

Definition

At $(x_1,y_1)$ : $\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$ .
At $\theta$ : $ax\sec\theta-by\csc\theta=a^2-b^2$ .
It meets the major axis at $x=\frac{(a^2-b^2)\cos\theta}{a}=ae^2\cos\theta$ .
The greatest distance of a normal from the centre is $a-b$ .

Normal at the parametric point

ax\sec\theta-by\csc\theta=a^2-b^2

Worked example

Find the normal to

\frac{x^2}{16}+\frac{y^2}{4}=1

(2,\sqrt3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q61Moderate

Example 3 · Conic Sections · Tangents, Normals and Chords of an Ellipse

If the maximum distance of normal to the ellipse

\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}= 1,b < 2

, from the origin is 1 , then the eccentricity of the ellipse is:

The normal has a MINUS between its terms

The tangent is

\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1

; the normal is

\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2

. The coordinates move to the denominators and the sign changes.

Drill 3 more on the normal to an ellipse

Concept 4 of 4: Chords: by midpoint, through a point, and at right angles at the centre

The chord with a given midpoint is

T=S_1

, as for every conic. When a line through a point

P

meets the ellipse at

A

and

B

, write the line as

x=x_0+r\cos\theta

y=y_0+r\sin\theta

: the two values of

r

are

PA

and

PB

, so their product comes straight from the quadratic. And two semi-diameters

OP

OQ

at right angles satisfy a fixed rule for their lengths.

Definition

Midpoint $(h,k)$ : $\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$ ; slope $-\frac{b^2h}{a^2k}$ .
Line through $P$ : substitute $(x_0+r\cos\theta,\ y_0+r\sin\theta)$ ; $PA\cdot PB=|r_1r_2|$ .
Perpendicular semi-diameters: $\frac{1}{OP^2}+\frac{1}{OQ^2}=\frac{1}{a^2}+\frac{1}{b^2}$ .
Midpoints of focal chords: put the focus into $T=S_1$ .

Chord with midpoint (h, k)

\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}

Worked example

Find the chord of

\frac{x^2}{16}+\frac{y^2}{9}=1

bisected at

(2,1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q134Moderate

Example 4 · Conic Sections · Tangents, Normals and Chords of an Ellipse

\alpha x + \beta y = 109

is the equation of the chord of the ellipse

\frac{x^{2}}{9}+\frac{y^{2}}{4}= 1

, whose mid point is

\left( \frac{5}{2},\frac{1}{2} \right)

, then

\alpha + \beta

is equal to

$T=S_1$ has $S_1$ on the right, not $1$

The chord with midpoint

(h,k)

ends in

\frac{h^2}{a^2}+\frac{k^2}{b^2}

. Writing

1

there gives the tangent-like line

T=0

, which passes through a different point.

Drill 9 more on chords: by midpoint, through a point, and at right angles at the centre

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 (4)

The tangent: point, parametric and slope forms
Tangency condition
$y=mx+c\ \text{touches}\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\iff c^2=a^2m^2+b^2$
Pairs of tangents, the chord of contact and the director circle
Director circle
$x^2+y^2=a^2+b^2$
The normal to an ellipse
Normal at the parametric point
$ax\sec\theta-by\csc\theta=a^2-b^2$
Chords: by midpoint, through a point, and at right angles at the centre
Chord with midpoint (h, k)
$\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$

Watch out for (4)

Shift the line before using $c^2=a^2m^2+b^2$
→ The tangent: point, parametric and slope forms
The director circle uses $a^2+b^2$ , not $a^2$
→ Pairs of tangents, the chord of contact and the director circle
The normal has a MINUS between its terms
→ The normal to an ellipse
$T=S_1$ has $S_1$ on the right, not $1$
→ Chords: by midpoint, through a point, and at right angles at the centre

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