Analytic Geometry – Problem 29: Find the standard equation of the ellipse

A central ellipse has equation $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\,(a>b>0)$. It is known that the tangent line to the ellipse at $P(4,3)$ is $3x+4y=24$.

Find the standard equation of the ellipse.

(1) For $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, the tangent line at $(x_0,y_0)$ on the ellipse is $$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1.$$ At $P(4,3)$, the tangent must be $$\frac{4x}{a^2}+\frac{3y}{b^2}=1.$$ (2) Since the given tangent line is $3x+4y=24$, dividing by $24$ gives $$\frac{x}{8}+\frac{y}{6}=1.$$ Match coefficients with $\frac{4x}{a^2}+\frac{3y}{b^2}=1$: $$\frac{4}{a^2}=\frac{1}{8}\Rightarrow a^2=32,\qquad \frac{3}{b^2}=\frac{1}{6}\Rightarrow b^2=18.$$ Thus the ellipse is $$\frac{x^2}{32}+\frac{y^2}{18}=1.$$ (3) Check: $\frac{4^2}{32}+\frac{3^2}{18}=\frac12+\frac12=1$, so $P$ lies on the ellipse. \]

The ellipse is $\dfrac{x^2}{32}+\dfrac{y^2}{18}=1$.