Let P be a parabola with vertex (2,3) and directrix 2 x+y=6.

Question

Let $P$ be a parabola with vertex $(2,3)$ and directrix $2 x+y=6$. Let an ellipse $$E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$$, of eccentricity $$\frac{1}{\sqrt{2}}$$ pass through the focus of the parabola $P$. Then, the square of ...

Accepted Answer

$\begin{gathered} =\frac{1}{2} \ y-3=\frac{1}{2}(x-2) \ \Rightarrow 2y-6=x-2 \ \Rightarrow 2y-x-4=0 \ 2x+y-6=0 \ 4x+2y-12=0 \ \alpha +1.6=4\Rightarrow \alpha =2.4 \ \beta +2.8=6\Rightarrow \beta =3.2 \end{gathered}$

Ellipse passes through $(2.4,3.2)$

$\Rightarrow \frac{{(\frac{24}{10})}^{2}}{{a}^{2}}+\frac{{(\frac{32}{10})}^{2}}{{b}^{2}}=1$ ..... (1)

Also $1-\frac{{b}^{2}}{{a}^{2}}=\frac{1}{2}=\frac{{b}^{2}}{{a}^{2}}=\frac{1}{2}$

$\Rightarrow {a}^{2}=2{b}^{2}$

Put in (1) $\Rightarrow {b}^{2}=\frac{328}{25}$

$\Rightarrow {(\frac{2{b}^{2}}{a})}^{2}=\frac{4{b}^{2}}{{a}^{2}}	imes {b}^{2}=4	imes \frac{1}{2}	imes \frac{328}{25}=\frac{656}{25}$

Solutions · 1