ધારો કે S અને S' નાભિઓ વાળા ઉપવલય frac{{{ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$

$e=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{25-16}}{5}=\frac{\sqrt{9}}{5}=\frac{3}{5}$

Vedclass · Accepted Answer

$12$

Vedclass · Answer

Equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$

$e=\frac{\sqrt{a^2-b^2}}{a}=\frac{\sqrt{25-16}}{5}=\frac{\sqrt{9}}{5}=\frac{3}{5}$

$\Rightarrow a e=5 	imes \frac{3}{5}=3$

Foci $F_1$ and $F_2$ are $(-3,0)$ and $(3,0)$ respectively.

$	ext { Area }=\frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \-3 & 0 & 1 \3 & 0 & 1 \end{array}ight|$ $=\frac{1}{2}[ x (0)- y (-3-3)+1(0)]$

$=\frac{-6 y }{2}=-3 y =3 	imes 4 	imes \sqrt{1-\frac{ x ^2}{25}} 	ext { since } y =4 \sqrt{1-\frac{ x ^2}{25}}$

$A=12 	imes \sqrt{1-\frac{x^2}{25}}$

for maximum value of $A$

$\frac{ d ( A )}{ dx }=0$

$\Rightarrow \frac{ d ( A )}{ dx }=12 	imes \frac{1}{2 \sqrt{1-\frac{ x ^2}{25}}} 	imes \frac{-2 x }{25}=0$

$\Rightarrow x =0$

So $A$ is $\max$ when $x =0$

$\Rightarrow A _{\max }=3 	imes 4 	imes \sqrt{1-0}=4 	imes 3 	imes \sqrt{1}=4 	imes 3=12 	ext { sq.units }$

$	herefore$ the maximum area is $12$ sq.units

Similar Questions

જો પરવલય $y^2 = x$ એ બિંદુ $\left( {\alpha ,\beta } \right)\,,\,\left( {\beta > 0} \right)$ અને ઉપવલય $x^2 + 2y^2 = 1$ આગળનો સ્પર્શક હોય તો $a$ =

ઉગમબિંદુ આગળ કેન્દ્રવાળા ઉપવલયની ઉત્કેન્દ્રતા $1/2$ છે. જો એક નિયામિકા $x = 4$ હોય તો ઉપવલયનું સમીકરણ :