Ellipse Questions

Solution : The equation x = acos$\theta$ & y = bsin$\theta$ together represent the parametric equation of ellipse ${x_1}^2\over a^2$ + ${y_1}^2\over b^2$ = 1, where $\theta$ is a parameter. Note that if P($\theta$) = (acos$\theta$, bsin$\theta$) is on the ellipse then ; Q($\theta$) = (acos$\theta$, bsin$\theta$) is on auxilliary circle. A circle described on […]

Solution : Let the equation of the ellipse be $x^2\over a^2$ + $y^2\over b^2$ = 1. Then, coordinates of the foci are $(\pm ae, 0)$. Therefore,  ae = 2 $\implies$  a = 4 We have $b^2$ = $a^2(1 – e^2)$ $\implies$ $b^2$ =12 Thus, the equation of the ellipse is $x^2\over 16$ + $y^2\over 12$

Solution : Let the equation of the required ellipse be $x^2\over a^2$ + $y^2\over b^2$ = 1                 ……….(i) Since the vertices of the ellipse are on y-axis. So, the coordinates of the vertices are $(0, \pm b)$. $\therefore$    b = 10 Now, $a^2$ = $b^2(1 – e^2)$ 

Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4 $\therefore$  mx – 2 = -1 $\implies$ m = $1\over 2$ Since $3x^2+4y^2$ = 12 or $x^2\over 4$ + $y^2\over 3$ = 1 Comparing this with $x^2\over a^2$ + $y^2\over b^2$ =

Solution : $\because$ Equation of ellipse is $9x^2 + 16y^2$ = 144 or $x^2\over 16$ + ${(y-3)}^2\over 9$ = 1 comparing this with $x^2\over a^2$ + $y^2\over b^2$ = 1 then we get $a^2$ = 16 and $b^2$ = 9 and comparing the line y = x + k with y = mx + c

Solution : Here S = (2, 3) & S’ is (-2, 3) and b = $\sqrt{5}$ $\implies$ SS’ = 4 = 2ae $\implies$ ae = 2 but $b^2$ = $a^2(1-e^2)$ $\implies$ 5 = $a^2$ – 4 $\implies$ a = 3 Hence the equation to major axis is y = 3. Centre of ellipse is midpoint