Ellipse

### 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 (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 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...

(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...
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{...