JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the tangent at a point \(P\) on the parabola \(y ^2=3 x\) is parallel to the line \(x+2 y=1\) and the tangents at the points \(Q\) and \(R\) on the ellipse \(\frac{x^2}{4}+\frac{y^2}{1}=1\) are perpendicular to the line \(x-y=2\), then the area of the triangle \(PQR\) is:
- A \(\frac{9}{\sqrt{5}}\)
- B \(5 \sqrt{3}\)
- C \(\frac{3}{2} \sqrt{5}\)
- D \(3 \sqrt{5}\)
Answer & Solution
Correct Answer
(D) \(3 \sqrt{5}\)
Step-by-step Solution
Detailed explanation
\(y^2=3 x\) Tangent \(P \left( x _1, y _1\right)\) is parallel to \(x +2 y =1\) Then slope at \(P =-\frac{1}{2}\) \(2 y \frac{ dy }{ dx }=3\) \(\Rightarrow \frac{ dy }{ dx }=\frac{3}{2 y }=-\frac{1}{2}\) \(\Rightarrow y_1=-3\) Coordinates of \(P (3,-3)\) Similarly…
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