Question

Find the equation to the tangent and normal at the ends of the latus rectam of the ellipse $9x^2+16y^2=144$.

Answer 1

$9x^2+16y^2=144\frac{9x^2}{144}+\frac{16y^2}{144}=1\frac{x^2}{16}+\frac{y^2}{9}=1a=4,b=3a>b$

So, the major axis of ellipse is $x$ axis.

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

Combining both we can say all the ends point are $(\pm ae,\pm \frac{b^2}{a})$

$e^2=1-\frac{b^2}{a^2}=1-\frac{9}{16}=\frac{7}{16}\Rightarrow e=\frac{\sqrt{7}}{4}ae=4.\frac{\sqrt{7}}{4}=\sqrt{7}\frac{b^2}{a}=\frac{9}{4}$

So the end point of latus rectum are $(\pm \sqrt{7},\pm \frac{9}{4})$

Equation of tangent

$\frac{x{x}^{\prime }}{a^2}+\frac{y{y}^{\prime }}{b^2}=1$

$\frac{\pm \sqrt{7}x}{16}+\frac{\pm \frac{9}{4}y}{9}=1\frac{\pm \sqrt{7}x}{16}\pm \frac{y}{4}=1\pm \sqrt{7}x\pm 4y=16$

Slope of tangent $m=\pm \frac{\sqrt{7}}{4}$

Let slope of normal $=m^{\prime }$

$m{m}^{\prime }=-1$

$\Rightarrow m^{\prime }=\mp \frac{4}{\sqrt{7}}$

Equation of normal

$y\pm \frac{9}{4}=\mp \frac{4}{\sqrt{7}}(x\pm \sqrt{7})\pm 4\sqrt{7}y\pm 9\sqrt{7}=\pm 16x\pm 16\sqrt{7}\pm 16x\mp 4\sqrt{7}y\pm 7\sqrt{7}=0\pm 4x\mp \sqrt{7}y=\frac{7\sqrt{7}}{4}$