Tangents are drawn to the ellipse 9 x 2 + 5 y 2 = 1 at ends of latusrectum. If the area of an quadrilateral be λ sq.units, then the value of λ /9 =

One end of latusrectum ( 2 , 3 5 ) Equation of tangent at ( 2 , 3 5 ) is 9 2 x + 3 y = 1 Area of quadrilateral = 4 × ( △ CPQ ) = 4 × 2 1 × 2 9 × 3 = 27 = λ ( Given ) ⇒ 9 λ = 3

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Tangents are drawn to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ at ends of latusrectum. If the area of an quadrilateral be $λ$ sq.units, then the value of $λ /9 =$

Answer:

Solution:

One end of latusrectum $(2, \frac{5}{3})$ Equation of tangent at $(2, \frac{5}{3})$ is $\frac{2 x}{9} + \frac{y}{3} = 1$

Area of quadrilateral $= 4 \times (△ CPQ) = 4 \times \frac{1}{2} \times \frac{9}{2} \times 3 = 27 = λ ($ Given $) \Rightarrow \frac{λ}{9} = 3$

Stream:JEESubject:MathematicsTopic:EllipseSubtopic:Tangent to ellipse

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