16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T4d8,{"@context":"https://schema.org","@type":"Quiz","name":"Mht Cet 2026 Ellipse PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/mht-cet/questions/2026/mathematics/ellipse/jt-mr31xy7r","educationalLevel":"Entrance exam preparation","about":"Mht Cet / Mathematics / Ellipse","hasPart":{"@type":"Question","name":"Mht Cet 2026 Ellipse PYQ with Solution","text":"If \\theta is the eccentric angle of a point on the ellipse \\dfrac{x^2}{25} + \\dfrac{y^2}{9} = 1 such that the distance of the point from the center is 5, then \\theta = .......","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"A"},"comment":{"@type":"Comment","text":"For the ellipse \\frac{x^2}{25}+\\frac{y^2}{9}=1 we have a^2=25,\\quad b^2=9 so a=5,\\quad b=3 If \\theta is the eccentric angle, then the point on the ellipse is written as x=a\\cos\\theta,\\quad y=b\\sin\\theta So here, x=5\\cos\\theta,\\quad y=3\\sin\\theta The distance of this point from the center (0,0) is given as 5. So, \\sqrt{x^2+y^2}=5 Squaring both sides, x^2+y^2=25 Substitute x=5\\cos\\theta and y=3\\sin\\theta: (5\\cos\\theta)^2+(3\\sin\\theta)^2=25 25\\cos^2\\theta+9\\sin^2\\theta=25 Now use \\cos^2\\theta=1-\\s..."}}}18:T415,

For the ellipse

$$ \frac{x^2}{25}+\frac{y^2}{9}=1 $$

we have

$$ a^2=25,\quad b^2=9 $$

$$ a=5,\quad b=3 $$

If $\theta$ is the eccentric angle, then the point on the ellipse is written as

$$ x=a\cos\theta,\quad y=b\sin\theta $$

So here,

Ellipse

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