Question

Let latus rectum of a parabola coincide with latus rectum of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ such that vertex of the parabola is $\left(x_1,0\right)$ where $x_1>3$. If area of the triangle formed by tangents to the parabola at end points of latus rectum and the $y$-axis is $S$, then find the value of $[\sqrt{S}]$, where $[k]$ denotes greatest integer less than or equal to $k$.

Answer 1

Answer: 6

$E:\frac{x^2}{25}+\frac{y^2}{16}=1\Rightarrow e^2=1-\frac{16}{25}=\frac{9}{25}$

$\Rightarrow e=\frac{3}{5}\Rightarrow F=(\pm 3,0)$
tangent at the extremeties of LR of parabola meet on the axis as well as directrix at $90^{\circ }$
$\therefore$ slope of each tangent is $m=\pm 1$
Now LR $=\frac{32}{5}=4a\Rightarrow a=\frac{8}{5}$
$\therefore$ distance between vertex \& focus is $\frac{8}{5}\&x_1\in (2,5)$
$\therefore$ parabola is as shown in the figure now tangent at $\left(3,\frac{16}{5}\right)$ with slope $m=-1$
$x+y=\frac{16}{5}+3;x=0,y=\frac{31}{5};y=0,x=\frac{31}{5}$
$\therefore$ Area of $\Delta =\frac{1}{2}\times 2\times \frac{31}{5}\times \frac{31}{5}$
$\therefore [\sqrt{S}]=6$.