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10-2. Parabola, Ellipse, Hyperbola
hard
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 .............. $\mathrm{sq. \,units}$
A
$27/4$
B
$9$
C
$27/2$
D
$27$
(IIT-2003)
Solution
(d) By symmetry the quadrilateral is a rhombus.
So area is four times the area of the right angled triangle formed by the tangent and axes in the Ist quadrant.
Now, $ae = \sqrt {{a^2} – {b^2}} $
$\Rightarrow ae = 2$
==> Tangent (in first quadrant) at end of latus rectum $\left( {2,\frac{5}{3}} \right)$ is
$\frac{2}{9}x + \frac{5}{3}\frac{y}{5} = 1$
$i.e.$, $\frac{x}{{9/2}} + \frac{y}{3} = 1$
Area $ = 4.\,\frac{1}{2}.\,\frac{9}{2}.3 = 27$ $sq.\ unit.$
Standard 11
Mathematics
