TS EAMCET · Maths · Application of Derivatives
The equation of the normal to the curve \(4 x^2+9 y^2=36\) at the point \(P\left(\frac{7 \pi}{4}\right)\) is
- A \(2 x-3 y-6 \sqrt{2}=0\)
- B \(2 x+3 y=0\)
- C \(3 \sqrt{2} x+2 \sqrt{2} y-5=0\)
- D \(3 \sqrt{2} x-2 \sqrt{2} y-13=0\)
Answer & Solution
Correct Answer
(C) \(3 \sqrt{2} x+2 \sqrt{2} y-5=0\)
Step-by-step Solution
Detailed explanation
Given curve \(4 x^2+9 y^2=36\) \[ \frac{x^2}{\left(\frac{36}{4}\right)}+\frac{y^2}{\left(\frac{36}{9}\right)}=1 \Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1 \] Equation of normal is \(a x \sec \theta-\) by \(\operatorname{cosec} \theta=a^2-b^2\)…
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