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Visual SolutionHard
Question
Let PP be a variable point on the ellipse x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 1 with foci SS and S′S'. If AA is the area of triangle PSS′PSS', then the maximum value of AA is:
Solution Path
Base = distance between foci = 6, max height = semi-minor axis = 4, area = 12.
01Question Setup
1/4
Find the maximum area of triangle PSS′PSS' where PP is on the ellipse x225+y216=1\frac{x^2}{25} + \frac{y^2}{16} = 1 and S,S′S, S' are foci.
Max area =;?= ;?
02Plot Ellipse
2/4
Ellipse with a=5a = 5, b=4b = 4, c=3c = 3. Foci at (±3,0)(\pm 3, 0). Base SS′=6SS' = 6.
SS′=6SS' = 6
03Maximum AreaKEY INSIGHT
3/4
Area =12×SS′×h= \frac{1}{2} \times SS' \times h. Height is maximized when PP is at end of minor axis: h=b=4h = b = 4.
Max height =b=4= b = 4
04Final Answer
4/4
Area =12×6×4=12= \frac{1}{2} \times 6 \times 4 = 12.
12\boxed{12}
Concepts from this question3 concepts unlocked
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Eccentricity and Conic Classification

EASY

The eccentricity e = c/a determines the type of conic: e < 1 gives an ellipse, e = 1 gives a parabola, and e > 1 gives a hyperbola. For a circle, e = 0.

e=cae = \frac{c}{a}

First step in most conic problems. JEE frequently asks you to identify or compare conics based on eccentricity values.

Conic identificationEccentricity comparisonLocus problems
Practice (12 Qs) →

Focal Chord Property

STANDARD

For a focal chord with endpoints P and Q on a conic, the semi-latus rectum l satisfies 1/SP + 1/SQ = 2/l, where S is the focus. This is a harmonic mean relation between SP and SQ.

1SP+1SQ=2l\frac{1}{SP} + \frac{1}{SQ} = \frac{2}{l}

Saves time in problems involving focal chords, latus rectum lengths, and minimum/maximum distance from the focus.

Focal chord lengthLatus rectumHarmonic mean problems
Practice (8 Qs) →

Parametric Representation of Conics

STANDARD

Parabola y² = 4ax uses (at², 2at). Ellipse x²/a² + y²/b² = 1 uses (a cos t, b sin t). Hyperbola x²/a² - y²/b² = 1 uses (a sec t, b tan t).

(at2, 2at),(acos⁡t, bsin⁡t),(asec⁡t, btan⁡t)(at^2,\, 2at), \quad (a\cos t,\, b\sin t), \quad (a\sec t,\, b\tan t)

Parametric form simplifies tangent, normal, and chord equations. Most JEE solutions use parametric coordinates for cleaner algebra.

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Practice (10 Qs) →