Locus of mid point of intercepts of tangents to a ellipse

Question

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!!!

Answer 1

Equation of tangent of ellipse is $$\frac{xx_1}{16}+\frac{yy_1}{9}=1 $$

Let's assume the midpoint of intercepts of the tangent to be $(h,k)$

The intercepts made by the tangent on the co-ordinate axes are $(\frac{16}{x_1},0)$ and $(0,\frac{9}{y_1})$ on x and y axes respectively.

Since $(h,k)$ will be the mid-point of the line segments joining the intercepts,

$h=\frac{16}{2x_1}=\frac{8}{x_1}$ and $k=\frac{9}{2y_1}$

So, $$x_1=\frac{8}{h}$$ and $$y_1=\frac{9}{2k}$$

But, since $x_1$ and $y_1$ are points on the ellipse, $$\frac{x_1^2}{16}+\frac{y_1^2}{9}=1 $$

$$\implies \frac{1}{16}(\frac{64}{h^2})+\frac{1}{9}(\frac{81}{4k^2})=1 $$

So, $$\frac{4}{h^2}+\frac{9}{4k^2}=1$$ $$\implies\frac{16}{h^2}+\frac{9}{k^2}=4$$

So, locus of the point $(h,k)$ is $$\frac{16}{x^2}+\frac{9}{y^2}=4$$ Refer to this image to see the graph of the function

Answer 2

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.