Find the equation of the ellipse, with major axis along the $x-$ axis and passing through the points $(4,\,3)$ and $(-1,\,4)$
Find the equation of the ellipse, with major axis along the $x-$ axis and passing through the points $(4,\,3)$ and $(-1,\,4)$
Solution The standard form of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 .$
since the points $(4,\,3)$ and $(-1,\,4)$ lie on the ellipse, we have
$\frac{16}{a^{2}}+\frac{9}{b^{2}}=1$ ............ $(1)$
and $\frac{1}{a^{2}}+\frac{16}{b^{2}}=1$ ......... $(2)$
Solving equations $(1)$ and $(2),$ we find that $a^{2}=\frac{247}{7}$ and $b^{2}=\frac{247}{15}$
Hence the required equation is
$\frac{x^{2}}{\left(\frac{247}{7}\right)}$ $+\frac{y^{2}}{\frac{247}{15}}=1,$ i.e., $7 x^{2}+15 y^{2}=247$
Similar Questions
A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$
A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$
- [IIT 2013]
Tangents at extremities of latus rectum of ellipse $3x^2 + 4y^2 = 12$ form a rhombus of area (in $sq.\ units$) -
Tangents at extremities of latus rectum of ellipse $3x^2 + 4y^2 = 12$ form a rhombus of area (in $sq.\ units$) -
An ellipse has $OB$ as semi minor axis, $F$ and $F'$ its foci and the angle $FBF'$ is a right angle. Then the eccentricity of the ellipse is
An ellipse has $OB$ as semi minor axis, $F$ and $F'$ its foci and the angle $FBF'$ is a right angle. Then the eccentricity of the ellipse is
- [AIEEE 2005]
The foci of the ellipse $25{(x + 1)^2} + 9{(y + 2)^2} = 225$ are at
The foci of the ellipse $25{(x + 1)^2} + 9{(y + 2)^2} = 225$ are at
The position of the point $(4, -3)$ with respect to the ellipse $2{x^2} + 5{y^2} = 20$ is
The position of the point $(4, -3)$ with respect to the ellipse $2{x^2} + 5{y^2} = 20$ is