A normal inclined at an angle of π4 to the x-axis of the ellipse x2a2+y2b2=1 is drawn. It meets the major and minor axes in P and Q respectively. If C is the centre of the ellipse then the area of the triangle CPQ is
A
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B
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C
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D
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Solution
The correct option is C Given ellipse is x2a2+y2b2=1 m=(dydx)(x1,y1)=−b2x1a2y1=−1 ( since the normal makes an angle π4 with x-axis) ⇒b2x1=a2y1→(1) Since P(x1,y1) lies the ellipse, we have b2x21+a2y21=a2b2→(2) From (1) & (2) P=(a2√a2+b2,b2√a2+b2) ∴ Equation of the normal at P is x – y = a2−b2√a2+b2→(3) Area of the △CPQ ie the area of the triangle made by the line (3) with the axes is (a2−b2)22(a2+b2)