7.1 Equation of tangent to a point/Point form

  Ellipse

It can be find out using $T=0$ i.e., equation of tangent to ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at point $({x_1},{y_1})$ is $$\frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} = 1$$

Question 7. Find the equation of tangent to the ellipse ${x^2} + 4{y^2} = 16$ at the end of latus rectum.

Solution: The equation of ellipse can be written as $$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{4} = 1$$

On comparing it with the standard equation of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, we get ${a^2} = 16$ and ${b^2} = 4$.

End points of latus rectum are $P( \ ae, \pm \frac{{{b^2}}}{a})$.

and $$\begin{equation} \begin{aligned} {e^2} = 1 - \frac{{{b^2}}}{{{a^2}}} \\ {e^2} = 1 - \frac{4}{{16}} \\ e = \frac{{\sqrt 3 }}{2} \\\end{aligned} \end{equation} $$

Therefore, the coordinates of end points of latus rectum are $$\begin{equation} \begin{aligned} P(4 \times \frac{{\sqrt 3 }}{2}, \pm \frac{4}{4}) \\ P(2\sqrt 3 , \pm 1) \\\end{aligned} \end{equation} $$

Now, the equation of tangents at point $P$ can be find out using $T=0$ i.e.,

$$\begin{equation} \begin{aligned} \frac{{x \times 2\sqrt 3 }}{{16}} + \frac{{y \times \pm 1}}{4} = 1 \\ \sqrt 3 x \pm 2y = 8 \\\end{aligned} \end{equation} $$