Let P be a parabola with vertex (2,3) and directrix 2x+y=6. Let an ellipse E: x^2 a^2+ y^2 b^2=1,a>b of eccentricity 1 √2 pass through the focus of…

Let $Z$Z be the foot of perpendicular from vertex to directrix of parabola,

Now, finding $Z$Z we get,

$$\Rightarrow \frac{x-2}{2}=\frac{y-3}{1}=-\frac{(4+3)-6}{4+1}$$⇒2x−2​=1y−3​=−4+1(4+3)−6​

$$\Rightarrow \frac{x-2}{2}=\frac{y-3}{1}=\frac{-1}{5}$$⇒2x−2​=1y−3​=5−1​

$$\Rightarrow \frac{x-2}{2}=\frac{-1}{5},\frac{y-3}{1}=\frac{-1}{5}$$⇒2x−2​=5−1​,1y−3​=5−1​

$$\Rightarrow x=\frac{12}{5},y=\frac{16}{5}$$⇒x=512​,y=516​

$$\Rightarrow Z\equiv (\frac{12}{5},\frac{16}{5})$$⇒Z≡(512​,516​)

Eccentricity of ellipse is given as $\frac{1}{\sqrt{2}}$2​1​.

Now, finding ${b}^{2}$b2 using eccentricity formula we get,

$$\Rightarrow {b}^{2}={a}^{2}(1-{e}^{2})=\frac{{a}^{2}}{2}$$⇒b2=a2(1−e2)=2a2​

So, equation of ellipse will be,

$\Rightarrow \frac{144}{25{a}^{2}}+\frac{256}{25\times \frac{{a}^{2}}{2}}=1$⇒25a2144​+25×2a2​256​=1 {as $(x,y)\equiv (\frac{12}{5},\frac{16}{5})$(x,y)≡(512​,516​) }

$$\Rightarrow \frac{144}{25{a}^{2}}+\frac{512}{25{a}^{2}}=1$$⇒25a2144​+25a2512​=1

$$\Rightarrow \frac{656}{25{a}^{2}}=1$$⇒25a2656​=1

$$\Rightarrow {a}^{2}=\frac{656}{25}$$⇒a2=25656​

$$\Rightarrow {b}^{2}=\frac{328}{25}$$⇒b2=25328​

Length of latus rectum is given by,

$L=\frac{2{b}^{2}}{a}$L=a2b2​.

$$\Rightarrow L=\frac{2\times \frac{328}{25}}{\frac{\sqrt{656}}{5}}$$⇒L=5656​​2×25328​​

$$\Rightarrow L=\frac{\sqrt{656}}{5}$$⇒L=5656​​

$$\Rightarrow {L}^{2}=\frac{656}{25}$$⇒L2=25656​