Let P(10,2 √15) be a point on the hyperbola x^2 a^2- y^2 ~b^2=1, whose foci are S and S^prime. If the length of its latus rectum is 8, then the…
JEE Main 2026 — Mathematics Coordinate Geometry
2026mcqmedium
Let P(10,215) be a point on the hyperbola a2x2−b2y2=1, whose foci are S and S′. If the length of its latus rectum is 8, then the square of the area of ΔPSS′ is equal to :
Official previous-year question
Held on 22 Jan 2026 · Verified 6 Jul 2026.
Options
A
900
B
4200
C
1462
D
2700
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Solution
For the hyperbola a2x2−b2y2=1, point P(10,215) gives a2100−b260=1.
The latus rectum length a2b2=8 gives b2=4a.
Substituting: a2100−a15=1, which yields a2+15a−100=0, so a=5 and b2=20.
Thus c2=45 and c=35.
The foci are S(±35,0).
Triangle PSS′ has base SS′=65 and height 215.
Area =21⋅65⋅215=303.
Therefore (Area)2=2700.
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