### Find the eccentricity of an ellipse whose latus rectum is one half of its minor axis.

### Find the equation of an ellipse whose eccentricity is

, the latus rectum is

, and the center is at the origin.

### Find the equation of an ellipse, the lengths of whose major and mirror axes are

and

units respectively.

### Find the equation of the ellipse which passes through the point

and having its foci at

.

### Find the equation of the ellipse with eccentricity

, foci on the y-axis, center at the origin and passing through the point

.

Given Eccentricity = \[\frac{3}{4}\] We know that Eccentricity = c/a Therefore,c=\[\frac{3}{4}\]a

### Find the equation of the ellipse with center at the origin, the major axis on the x-axis and passing through the points

and

.

Given: Center is at the origin and Major axis is along x – axis So, Equation of ellipse is of the form \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Given that ellipse passing...

### Find the equation of the ellipse whose foci are at

and

Given: Coordinates of foci = \[\left( \mathbf{0},\text{ }\pm \mathbf{4} \right)\] …(i) We know that, Coordinates of foci = \[\left( 0,\text{ }\pm c \right)\] …(ii) The coordinates of the foci are...

### Find the equation of the ellipse whose foci are at

and e=1/2

Let the equation of the required ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Given: Coordinates of foci = \[\left( \pm 1,\text{ }0 \right)\] …(i) We know that,...

### Find the equation of the ellipse whose foci are

and the eccentricity is

Let the equation of the required ellipse be Given: Coordinates of foci = \[\left( \pm 2,\text{ }0 \right)\]…(iii) We know that, Coordinates of foci = \[\left( \pm c,\text{ }0 \right)\]…(iv) ∴ From...

### Find the equation of the ellipse the ends of whose major and minor axes are

and

respectively.

Given: Ends of Major Axis = \[\left( \pm \mathbf{4},\text{ }\mathbf{0} \right)\] and Ends of Minor Axis = \[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\] Here, we can see that the major axis is...

### Find the equation of the ellipse whose vertices are the

and foci at

.

Given: Vertices = \[(0,\pm 4)\] …(i) The vertices are of the form = (0, ±a) …(ii) Hence, the major axis is along y – axis ∴ From eq. (i) and (ii), we get \[\begin{array}{*{35}{l}} a\text{ }=\text{...

### Find the equation of the ellipse whose vertices are at

and foci at

.

Given: Vertices = \[\left( \pm \mathbf{6},\text{ }\mathbf{0} \right)\] …(i) The vertices are of the form = \[\left( \pm a,\text{ }0 \right)\] …(ii) Hence, the major axis is along x – axis ∴ From eq....

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: \[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: \[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}\] Divide by \[16\] to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}\]…(i) Divide by \[18\] to both the sides, we get...

### Find the (v) length of the latus rectum of each of the following ellipses.

Answer :

Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing eq. (i)...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing eq. (i)...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]…(i) Since, \[4\text{ }<\text{ }25\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii)...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]…(i) Since, \[4\text{ }<\text{ }25\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii)...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since,...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since,...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since, ...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: \[\frac{{{x}^{2}}}{49}+\frac{{{y}^{2}}}{36}=1\]…(i) Since, \[49\text{ }>\text{ }36\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] …(ii)...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: \[\frac{{{x}^{2}}}{49}+\frac{{{y}^{2}}}{36}=1\]…(i) Since, \[49\text{ }>\text{ }36\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] …(ii)...

### Find the(v) length of the latus rectum of each of the following ellipses.

Given: \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Since, \[25>9\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(ii)...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Since, \[25>9\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(ii)...

### Find the eccentricity of an ellipse whose latus – rectum is (i) Half of its minor axis (ii) Half of its major axis

Given: We need to find the eccentricity of an ellipse. (i) If latus – rectum is half of its minor axis We know that the length of the semi – minor axis is \[b\]and the length of the latus – rectum...

### Find the equation of the ellipse whose centre is (-2, 3) and whose semi – axis are 3 and 2 when the major axis is (i) parallel to x – axis (ii) parallel to the y – axis.

Given: Centre \[=\text{ }\left( -2,\text{ }3 \right)\] Semi – axis are \[3\text{ }and\text{ }2\] (i) When major axis is parallel to x-axis Now let us find the equation to the ellipse. We know that...

### Find the equation of the ellipse in the standard form whose minor axis is equal to the distance between foci and whose latus – rectum is 10.

Given: Minor axis is equal to the distance between foci and whose latus – rectum is \[10\] Now let us find the equation to the ellipse. We know that the equation of the ellipse whose axes are x and...

### Find the equation of the ellipse whose foci are (4, 0) and (- 4, 0), eccentricity = 1/3.

Given: Foci are \[\left( 4,\text{ }0 \right)\text{ }\left( -\text{ }4,\text{ }0 \right)\] Eccentricity \[=\text{ }1/3\] Now let us find the equation to the ellipse. We know that the equation of the...

### Find the equation of the ellipse in the following cases: The ellipse passes through (1, 4) and (- 6, 1)

The ellipse passes through \[\left( 1,\text{ }4 \right)\text{ }and\text{ }\left( -\text{ }6,\text{ }1 \right)\] Given: The points \[\left( 1,\text{ }4 \right)\text{ }and\text{ }\left( -\text{...

### Find the equation of the ellipse in the following cases: (i) eccentricity e = ½ and semi – major axis = 4 (ii) eccentricity e = ½ and major axis = 12

(i) \[\mathbf{eccentricity}\text{ }\mathbf{e}\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}~\]and \[\mathbf{semi}\text{ }\text{ }\mathbf{major}\text{ }\mathbf{axis}\text{ }=\text{ }\mathbf{4}\]...

### Find the equation of the ellipse in the following cases: (i) eccentricity e = ½ and foci (± 2, 0) (ii) eccentricity e = 2/3 and length of latus – rectum = 5

(i) \[Eccentricity\text{ }e\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}~\] and \[foci\text{ }\left( \pm \text{ }2,\text{ }0 \right)\] Given: Eccentricity \[e\text{ }=\text{ }{\scriptscriptstyle...

### Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (-3, 1) and has eccentricity √(2/5).

Given: The point \[\left( -3,\text{ }1 \right)\] Eccentricity \[=\text{ }\surd \left( 2/5 \right)\] Now let us find the equation to the ellipse. We know that the equation of the ellipse whose axes...

### Find the eccentricity, coordinates of foci, length of the latus – rectum of the following ellipse: 9x^2 + 25y^2 = 225

\[9{{x}^{2}}~+\text{ }25{{y}^{2}}~=\text{ }225\] Given: The equation of ellipse \[=>\text{ }9{{x}^{2}}~+\text{ }25{{y}^{2}}~=\text{ }225\] This equation can be expressed as By using the formula,...

### Find the eccentricity, coordinates of foci, length of the latus – rectum of the following ellipse: (i) 4×2 + 3y2 = 1 (ii) 25×2 + 16y2 = 1600

(i) \[4{{x}^{2}}~+\text{ }3{{y}^{2}}~=\text{ }1\] Given: The equation of ellipse \[=>\text{ }4{{x}^{2}}~+\text{ }3{{y}^{2}}~=\text{ }1\] This equation can be expressed as By using the formula,...

### Find the eccentricity, coordinates of foci, length of the latus – rectum of the following ellipse: (i) 4×2 + 9y2 = 1 (ii) 5×2 + 4y2 = 1

(i) \[4{{x}^{2}}~+\text{ }9{{y}^{2}}~=\text{ }1\] Given: The equation of ellipse \[=>\text{ }4{{x}^{2}}~+\text{ }9{{y}^{2}}~=\text{ }1\] This equation can be expressed as By using the formula,...

### Find the equation of the ellipse in the following cases: (i) focus is (- 2, 3), directrix is 2x + 3y + 4 = 0 and e = 4/5. (ii) focus is (1, 2), directrix is 3x + 4y – 7 = 0 and e = ½.

(i) \[focus\text{ }is\text{ }\left( -\text{ }2,\text{ }3 \right),\] \[directrix\text{ }is-\text{ }2x\text{ }+\text{ }3y\text{ }+\text{ }4\text{ }=\text{ }0\] and \[e\text{ }=\text{ }4/5\] Focus is...

### Find the equation of the ellipse in the following cases: (i) focus is (0, 1), directrix is x + y = 0 and e = ½. (ii) focus is (- 1, 1), directrix is x – y + 3 = 0 and e = ½.

(i) \[focus\text{ }is\text{ }\left( 0,\text{ }1 \right),\] \[directrix\text{ }is\text{ }x\text{ }+\text{ }y\text{ }=\text{ }0\]and \[e\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\] Given: Focus...

### Find the equation of the ellipse whose focus is (1, -2), the directrix 3x – 2y + 5 = 0 and eccentricity equal to 1/2.

Given: \[Focus\text{ }=\text{ }\left( 1,\text{ }-2 \right)\] \[Directrix\text{ }=\text{ }3x\text{ }-\text{ }2y\text{ }+\text{ }5\text{ }=\text{ }0\] \[Eccentricity\text{ }=\text{...