Find the equation of the ellipse

Question

An ellipse with centre at $(4,3)$ touches $x$-axis at $(0,0)$. If the slope of the major axis of ellipse is 1, then find the equation of the ellipse?

Answer 1

An ellipse centred at $(4,3)$ with major and minor axes parallel to the $x-$ and $y-$ axes respectively has the following equation: $$\frac {(x-4)^2}{a^2}+\frac{(y-3)^2}{b^2}=1\qquad (a>b>0)\qquad\cdots(1)$$ Applying the rotation matrix for $\theta=-\pi/4$ and simplifying: $$\frac{(x+y-7)^2}{2a^2}+\frac{(x-y-1)^2}{2b^2}=1\qquad\cdots(2)$$ Differentiating w.r.t. $x$ and rearranging: $$\frac{dy}{dx}\left[\frac{x+y-7}{a^2}-\frac{x-y-1}{b^2}\right]=-\frac{x+y-7}{a^2}-\frac{x-y-1}{b^2}\qquad\cdots (3)$$ Ellipse passes through $(0,0)$. Substituting in $(1)$ and rearranging: $$ \frac{49}{2a^2}+\frac 1{2b^2}=1\qquad\cdots(4)$$ As $(0,0)$, ellipse touches the $y-$axis, i.e $\frac{dy}{dx}=\infty$. From $(3)$, $$ a^2=7b^2\qquad\cdots(5)$$ From $(4),(5)$, $$a=2\sqrt{7}, b=2$$ Substituting in $(1)$ gives equation of the tilted ellipse: $$\frac{(x+y-7)^2}{56}+\frac{(x-y-1)^2}{8}=1\qquad\blacksquare$$ This is the same as $$2x^2+2y^2-3xy-7x=0$$