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

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

$\Rightarrow e =\frac{3}{5} \Rightarrow F =( \pm 3,0)$
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 $=\frac{32}{5}=4 a \Rightarrow a =\frac{8}{5}$
$\therefore$ 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 $\Delta=\frac{1}{2} \times 2 \times \frac{31}{5} \times \frac{31}{5}$
$\therefore [\sqrt{ S }]=6$.