The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse x 2 9 + y 2 5 = 1 , is [2003]

Question

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ , is [2003]

Smart Achievers · Accepted Answer

(4)

$Given equation of ellipse is \frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$

$∴$ $a^{2} = 9, b^{2} = 5 \Rightarrow e = \sqrt{1 - \frac{5}{9}} = \frac{2}{3}$

$∴$ $End point of latus rectum in first quadrant is L (2, \frac{5}{3})$

$Equation of tangent at L is \frac{2 x}{9} + \frac{y}{3} = 1,$ $which meets x-axis at A (\frac{9}{2}, 0) and y-axis at B (0, 3)$

$∴$ $Area of ∆ O A B = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{27}{4}$

$Now by symmetry area of quadrilateral A B C D$

$= 4 \times Area (∆ O A B) = 4 \times \frac{27}{4} = 27 sq. units.$

Solution

Q.
The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ , is [2003]

1 $\frac{27}{4} sq. units$

2 $9 sq. units$

3 $\frac{27}{2} sq. units$

4 $27 sq. units$

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Solution

Q. The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse x29+y25=1, is [2003]

1 274 sq. units

2 9 sq. units

3 272 sq. units