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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 391-392 of 392 questions
391 Easy
391 of 392
Application +4 | -1
If the eccentricities of the two ellipse x2169+y225=1\frac{x^2}{169}+\frac{y^2}{25}=1 and x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 are equal, then the value ab\frac{a}{b}, is
View answer and explanation
Correct option: C
135\frac{13}{5}
Option C (135\frac{13}{5}) is correct because it is derived from equating the eccentricities of the two ellipses and solving for the ratio ab\frac{a}{b}.
392 Easy
392 of 392
Application +4 | -1
The equation of the normal to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at the point (acos⁡θ,bsin⁡θ)(a \cos \theta, b \sin \theta) is
View answer and explanation
Correct option: C
axcos⁡θ−bysin⁡θ=a2−b2\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2
Option C (axcos⁡θ−bysin⁡θ=a2−b2\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2) is correct because it is the standard form equation of the normal to the ellipse at the given point.
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