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MHT-CET Maths · Conic Sections

Ellipse and Hyperbola — Eccentricity, Tangents and Orthogonal Curves

The eccentricity of an ellipse and a hyperbola and what fixes it, and the tangent condition c² = a²m² ± b², used for tangents of a given slope, for the area a tangent cuts off, and for curves that cross at right angles.

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Why this matters

10 PYQs, four HARD. Five ask for an eccentricity or an equation — an ellipse after completing the square, a hyperbola through two points, a curve given parametrically, a hyperbola sharing an ellipse's foci. Five are tangents: a tangent of given slope and its intercepts, and two curves that cut at right angles. Two cards.

Concept 1 of 2: Eccentricity and Foci of the Ellipse and Hyperbola

Eccentricity measures how stretched a conic is. For an ellipse the foci are inside, at distance ae from the centre, and e < 1; b² = a²(1 − e²). For a hyperbola they are outside and e > 1; b² = a²(e² − 1). In both, a belongs to the axis that carries the foci. When the larger denominator sits under y², the ellipse's major axis is vertical, and the formula uses that larger number as a².

Definition

  • Ellipse x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1, a>ba > b: e=1−b2a2e = \sqrt{1 - \dfrac{b^2}{a^2}}, foci (±ae,0)(\pm ae, 0).
  • If the larger denominator is under y2y^2, swap roles: e=1−smallerlargere = \sqrt{1 - \dfrac{\text{smaller}}{\text{larger}}}.
  • Hyperbola x2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1: e=1+b2a2e = \sqrt{1 + \dfrac{b^2}{a^2}}, foci (±ae,0)(\pm ae, 0).
  • A curve through given points: substitute each point to get a2a^2 and b2b^2.
  • Parametric x=p(cos⁡t+sin⁡t)x = p(\cos t + \sin t), y=q(cos⁡t−sin⁡t)y = q(\cos t - \sin t): x2p2+y2q2=2\dfrac{x^2}{p^2} + \dfrac{y^2}{q^2} = 2, an ellipse.
  • Foci subtending a right angle at an end of the minor axis: b=aeb = ae, so e=12e = \dfrac{1}{\sqrt2}.

Eccentricity

ellipse: b2=a2(1−e2)hyperbola: b2=a2(e2−1)\text{ellipse: } b^2 = a^2(1 - e^2) \qquad \text{hyperbola: } b^2 = a^2(e^2 - 1)

Worked example

Find the eccentricity of 4x2+3y2−8x=84x^2 + 3y^2 - 8x = 8.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q146Moderate

Example 1 · Conic Sections · Ellipse and Hyperbola

The eccentricity of the ellipse 9x2+5y2−30y=09x^{2}+ 5y^{2}- 30y= 0 is

Dividing by the wrong denominator

In x²/5 + (y − 3)²/9 = 1 the major axis is vertical. e = √(1 − 5/9) = 2/3. Always put the smaller denominator on top; 1 − 9/5 is negative.

Using the ellipse relation for a hyperbola

An ellipse has b² = a²(1 − e²); a hyperbola has b² = a²(e² − 1). A hyperbola's e is always greater than 1, so any option below 1 is out.

Forgetting the 2 in the parametric form

(cos t + sin t)² + (cos t − sin t)² = 2, not 1. So x = 3(cos t + sin t), y = 4(cos t − sin t) is x²/18 + y²/32 = 1, whose e is √7/4.
Drill 4 more on eccentricity and foci of the ellipse and hyperbola

Concept 2 of 2: Tangents of a Given Slope, and Curves That Cut at Right Angles

As with the parabola, y = mx + c touches an ellipse or hyperbola when the meeting-point quadratic has equal roots. That gives c² = a²m² + b² for the ellipse and c² = a²m² − b² for the hyperbola. Two curves cut orthogonally when their tangents at a common point are perpendicular. For two central conics this has a shortcut: they must have the same foci.

Definition

  • Ellipse: y=mx±a2m2+b2y = mx \pm \sqrt{a^2m^2 + b^2}. Hyperbola: y=mx±a2m2−b2y = mx \pm \sqrt{a^2m^2 - b^2}.
  • Tangent at a point: xx1a2±yy1b2=1\dfrac{xx_1}{a^2} \pm \dfrac{yy_1}{b^2} = 1; at (asec⁡θ,btan⁡θ)(a\sec\theta, b\tan\theta) on the hyperbola, xsec⁡θa−ytan⁡θb=1\dfrac{x\sec\theta}{a} - \dfrac{y\tan\theta}{b} = 1, slope basin⁡θ\dfrac{b}{a\sin\theta}.
  • Equal intercepts: slope −1-1. A tangent's intercepts and the origin form a triangle of area 12∣x-int⋅y-int∣\frac12|x\text{-int}\cdot y\text{-int}|.
  • Orthogonal curves: at the common point, m1m2=−1m_1m_2 = -1. For A1x2+B1y2=1A_1x^2 + B_1y^2 = 1 and A2x2+B2y2=1A_2x^2 + B_2y^2 = 1: 1A1−1B1=1A2−1B2\dfrac{1}{A_1} - \dfrac{1}{B_1} = \dfrac{1}{A_2} - \dfrac{1}{B_2} (confocal).

Tangent of slope m

ellipse: c2=a2m2+b2hyperbola: c2=a2m2−b2\text{ellipse: } c^2 = a^2m^2 + b^2 \qquad \text{hyperbola: } c^2 = a^2m^2 - b^2

Worked example

Find the tangents to x29+y24=1\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 that are perpendicular to y=xy = x.
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q141Hard

Example 2 · Conic Sections · Ellipse and Hyperbola

The tangent to the ellipse 9x2+16y2=2889x^{2}+ 16y^{2}= 288 making equal intercepts on the co-ordinate axes intersects the X -axis and the Y -axis in the points A and B respectively. Then A(△OAB)=A( \bigtriangleup OAB) = (where O is origin)

Using the ellipse sign for a hyperbola

The ellipse adds b², the hyperbola subtracts it. With x²/20 − y²/5 = 1 and m = 3/4, c² = 20(9/16) − 5 = 25/4, so c = ±5/2.

Not writing the ellipse in standard form first

9x² + 16y² = 288 is x²/32 + y²/18 = 1. The tangent condition uses 32 and 18, not 9 and 16.

Slope of a perpendicular line

A tangent perpendicular to 4x + 3y = 7 has slope 3/4, the negative reciprocal of −4/3. Using −4/3 gives a tangent parallel to the line.
Drill 4 more on tangents of a given slope, and curves that cut at right angles

Summary — formulas & gotchas at a glance

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