AP EAMCET · Maths · Ellipse
The area (in sq. units) of the triangle formed by the tangent and the normal at the point \(\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)\) to the curve \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and the \(X\)-axis is
- A \(\frac{a}{b}\left(a^2+b^2\right)\)
- B 4ab
- C \(\frac{b}{4 a}\left(a^2+b^2\right)\)
- D 2ab
Answer & Solution
Correct Answer
(C) \(\frac{b}{4 a}\left(a^2+b^2\right)\)
Step-by-step Solution
Detailed explanation
Given curve, \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] Differentiate w.r.t \(x\), we get \(\Rightarrow \quad \frac{2 x}{a^2}+\frac{2 y \cdot y^{\prime}}{b^2}=0 \Rightarrow y^{\prime}=\frac{-b^2 x}{y a^2}\) At…
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