The equation of the ellipse with axes along the x-axis and the y-axis, which passes through the points P(4, 3) and Q (6, 2) is

axes ary $x$ auls, jants the equation of elliple will be,

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

If it goes through $(4,3)$

$\begin{array}{l} \rightarrow \frac{(4)^{2}}{a^{2}}+\frac{3^{2}}{b^{2}}-5 \\ \Rightarrow \frac{16}{a^{2}}+\frac{9}{b^{2}}=1 \end{array}$

if it goes through $(6,2)$

$\begin{array}{l} \rightarrow \frac{(6)^{2}}{a^{2}}+\frac{2^{2}}{b^{2}}=1 \\ \Rightarrow \frac{36}{a^{2}}+\frac{4}{b^{2}}=1 \\ \Rightarrow \frac{36}{a^{2}}=1-\frac{4}{b^{2}} \\ \Rightarrow x^{2}=\frac{36}{\left(\frac{b^{2}-4}{b^{2}}\right)}=\frac{36 b^{2}}{b^{2}-4} \end{array}$

Ler, put the raluen, $a^{2}$ in - (1)

$\begin{aligned} & 16 \times \frac{b^{2}-4}{^{36 b^{2}}}+\frac{9}{b^{2}}=1 \\ \Rightarrow & \frac{9 b^{2}-16+81}{9 b^{2}}=1 . \end{aligned}$

$\begin{array}{l} \Rightarrow \quad 4 b^{2}+65=9 b^{2} \\ \Rightarrow \quad 65=5 b^{2} \\ \therefore b^{2}=13 \\ \therefore \quad a^{2}=\frac{36 \times 13}{13-4}=\frac{36 \times 13}{9}=52 \end{array}$

equation will be

$\frac{x^{2}}{52}+\frac{y^{2}}{13}=1$