# Ellipse Questions

## What is the parametric equation of ellipse ?

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 …

## The foci of an ellipse are $$(\pm 2, 0)$$ and its eccentricity is 1/2, find its equation.

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 $$x^2\over 16$$ + $$y^2\over 12$$ …

## Find the equation of the ellipse whose axes are along the coordinate axes, vertices are $$(0, \pm 10)$$ and eccentricity e = 4/5.

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, $$a^2$$ = $$b^2(1 – e^2)$$  …

## If the foci of a hyperbola are foci of the ellipse $$x^2\over 25$$ + $$y^2\over 9$$ = 1. If the eccentricity of the hyperbola be 2, then its equation is :

Solution : For ellipse e = $$4\over 5$$, so foci = ($$\pm$$4, 0) for hyperbola e = 2, so a = $$ae\over e$$ = $$4\over 2$$ = 2, b = $$2\sqrt{4-1}$$ = 2$$\sqrt{3}$$ Hence the equation of the hyperbola is $$x^2\over 4$$ – $$y^2\over 12$$ = 1 Similar Questions Find the equation of the ellipse …

## Find the equation of the tangents to the ellipse $$3x^2+4y^2$$ = 12 which are perpendicular to the line y + 2x = 4.

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 $$x^2\over a^2$$ + $$y^2\over b^2$$ = …

## For what value of k does the line y = x + k touches the ellipse $$9x^2 + 16y^2$$ = 144.

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 …

## Find the equation of ellipse whose foci are (2, 3), (-2, 3) and whose semi major axis is of length $$\sqrt{5}$$.

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 …