Question

Find the area of the quadrilateral formed by the tangents at the end points of latus ractum to the ellipse
$\frac{ {x}^{2} }{9} + \frac{ {y}^{2} }{5} = 1$


✔️✔️Proper solution needed ✔️✔️​

Answer 1

Answer:

Step-by-step explanation:

❣Holla user❣

I think these pic would helps u^_^

Answer 2

ANSWER

27 sq. units.

Step By step Explanation

plz refer to attachment for the diagram.

Equation of tangent at (ae,$\dfrac{b^{2} }{a}$) is >

$\dfrac{x.ae}{a^{2} } + \dfrac{\dfrac{y.b^{2} }{a} }{b^{2}}$ = 1

=>$\dfrac {ex}{a} + \dfrac {y}{a}$ =1

i.e we get > ex + y = a.

Tangent internet x and y axis at P and Q.

P =($\dfrac{a}{e}$,0)

Q = (0,a)

Now, we know that=>

Area of quadrilateral = 4 * Area if triangle POQ.

=> Area of Quadrilateral =$4 \times \dfrac{1}{2} \times\dfrac {a}{e}\times a$

=> Area of Quadrilateral = $\dfrac{4a^{2} }{2e}$ or $\dfrac{2a^{2} }{e}$

Here , a = √9 and b = √5

e = $\sqrt(1 - \dfrac{b^{2} }{a^{2} })$

=>e = $\sqrt(1 - \dfrac{5}{9})$

=> e = $( \dfrac{2}{3} )$

Hence the area of the quadrilateral Will be >>

$\dfrac{2a^{2} }{e}$

=> $\dfrac{2\times 3^{2} }{\dfrac{2}{3} }$

=> 27 sq. units.

Hence the area if quadrilateral is 27 sq units