Find the equation of the locus of mid points of the portion of the tangent to the ellipse $$\frac{x^2}{16} + \frac{y^2}{9} = 1$$ intercepted between the axes
The answer is $$\frac{16}{x^2} + \frac{9}{y^2} = 4$$
Honestly speaking my mind is boggled by this question. I don't understand a bit. I know how to calculate the tangent to an ellipse through a point.
""If the point is $P(x_1,y_1)$ then the equation of tangent is $\frac{xx_1}{16} + \frac{yy_1}{9}=1$ for the given e equation""
But to find such a locus is just out of my mind. Please help!!!
Best Answer
The ellipse has equation $$\frac{x^2}{16}+\frac{y^2}{9}=1$$ Applying $\frac{d}{dx}$ you get
\begin{align} \frac x8 + \frac{2y}{9} \frac{dy}{dx}&=0 \\ \implies \frac{dy}{dx} & = -\frac{9x}{16y} \end{align}
i.e. at a point $(x,y)$ on the ellipse, the tangent has gradient $-\frac{9x}{16y}$.
Parameterize the ellipse as $(x(t),y(t))=(4cost,3sint)$ where $0≤t<2 \pi$. Then at such a point, the tangent has gradient $-\frac{9(4cost)}{16(3sint)}=-\frac 34 cot(t)$, so that the equation of the tangent is $$y=-\frac 34 x cot(t)+3csc(t)$$
This line intersects the $x$ and $y$ axes at the points $(4sec(t),0)$ and $(0,3csc(t))$respectively. The midpoint between between these two points would be $(2sec(t),1.5csc(t))$, and recall that $0≤t<2\pi$.
So, in parametric form, the locus of midpoints of tangents is
\begin{align} & x=\frac{2}{cos(t)} \\ & y=\frac{3}{2sin(t)} \\ \end{align}
i.e. $\frac 1x = \frac 12 cos(t)$ and $\frac 1y = \frac 23 sin(t)$, and we observe that they satisfy $4 \frac {1}{x^2} + \frac 94 \frac {1}{y^2} = 1$, hence $$\frac{16}{x^2}+\frac{9}{y^2}=4$$ as required.