JEE Main 2007 — Application of Derivatives Question with Solution
JEE Main 2007
Question
The normal to a curve at meets the -axis at . If the distance of from the origin is twice the abscissa of , then the curve is a
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Show full solutionCorrect option: A
Correct answer
Aellipse
Step-by-step explanation
Equation of normal is
$\begin{aligned}
& \Rightarrow G \equiv\left(x+y \frac{d y}{d x}, 0\right) \\
& \left|x+y \frac{d y}{d x}\right|=|2 x| \\
& \Rightarrow y \frac{d y}{d x}=x \text { or } y \frac{d y}{d x}=-3 x \\
& y d y=x d x \text { or } y d y=-3 x d x \\
& \frac{y^2}{2}=\frac{x^2}{2}+c \text { or } \frac{y^2}{2}=-\frac{3 x^2}{2}+c \\
& x^2-y^2=-2 c \text { or } 3 x^2+y^2=2 c.
\end{aligned}$
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This is a previous-year question from JEE Main 2007, covering the Application of Derivatives chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.