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:T501,{"@context":"https://schema.org","@type":"Quiz","name":"Ts Eamcet 2023 Ellipse PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/ts-eamcet/questions/2023/mathematics/ellipse/mdfbar14","educationalLevel":"Entrance exam preparation","about":"Ts Eamcet / Mathematics / Ellipse","hasPart":{"@type":"Question","name":"Ts Eamcet 2023 Ellipse PYQ with Solution","text":"Tangents are drawn to the ellipse \\frac{x^2}{9}+\\frac{y^2}{5}=1 at all the ends of its latus recta. The area of the quadrilateral, so formed (in sq units) is","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"A"},"comment":{"@type":"Comment","text":"The equation of ellipse \\frac{x^2}{9}+\\frac{y^2}{5}=1 So, e=\\sqrt{1-\\frac{5}{9}}=\\sqrt{\\frac{4}{9}}=\\frac{2}{3} So, a e=2 Now, equation of tangent at \\left(2, \\frac{5}{3}\\right) is given as T:\\left(\\frac{2}{9}\\right) x+\\frac{y}{3}=1 \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...(i) and equation of tangent at \\left(2,-\\frac{5}{3}\\right) is given as T^1:\\left(\\frac{2}{9}\\right) x+\\left(-\\frac{y}{3}\\right)=1 \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...(i..."}}}18:T5d9,

The equation of ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$

So, $e=\sqrt{1-\frac{5}{9}}=\sqrt{\frac{4}{9}}=\frac{2}{3}$

So, $a e=2$

Now, equation of tangent at $\left(2, \frac{5}{3}\right)$ is

given as

$$ T:\left(\

Ellipse

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