Mathematics - Ellipse Question with Solution | TestHub

Let $(h,k)$ be the midpoint of chord, then equation of chord is $\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$

a) Homogeniring we get equation of pair OA and OB

$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right){\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}\right)}^2={\left(\frac{xh}{a^2}+\frac{yk}{b^2}\right)}^2$

Pair of lines OA and OB are perpendicular ⇒ Coefficient of $x^2+$ Coefficient of $y^2=0$ $\Rightarrow {\left(\frac{{\mathrm{h}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{k}}^2}{{\mathrm{b}}^2}\right)}^2\left(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}\right)-\left(\frac{{\mathrm{h}}^2}{{\mathrm{a}}^4}+\frac{{\mathrm{k}}^2}{{\mathrm{b}}^4}\right)=0$

Locus is $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=\frac{x^2}{a^4}+\frac{y^2}{b^4}$

b) Put $(\mathrm{h}+\mathrm{r}\cos \theta ,\mathrm{k}+\mathrm{r}\sin \theta )$ to equation of ellipse $\frac{{\mathrm{h}}^2+{\mathrm{r}}^2{\cos }^2\theta +(2\mathrm{h}\cos \theta )\mathrm{r}}{{\mathrm{a}}^2}+\frac{{\mathrm{k}}^2+{\mathrm{r}}^2{\sin }^2\theta +(2\mathrm{k}\sin \theta )\mathrm{r}}{{\mathrm{b}}^2}=1$ $\Rightarrow {\mathrm{r}}^2\left(\frac{{\cos }^2\theta }{{\mathrm{a}}^2}+\frac{{\sin }^2\theta }{\mathrm{b}}\right)+\left(\frac{2\mathrm{h}\cos \theta }{{\mathrm{a}}^2}+\frac{2\mathrm{k}\sin \theta }{{\mathrm{b}}^2}\right)\mathrm{r}+\left(\frac{{\mathrm{h}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{k}}^2}{{\mathrm{b}}^2}-1\right)=0$ $\Rightarrow \frac{2\mathrm{h}\cos \theta }{{\mathrm{a}}^2}+\frac{2\mathrm{k}\sin \theta }{{\mathrm{b}}^2}=0\Rightarrow \tan \theta =-\frac{{\mathrm{b}}^2\mathrm{h}}{{\mathrm{a}}^2\mathrm{k}}$ and $\frac{\left(\frac{{\mathrm{h}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{k}}^2}{{\mathrm{b}}^2}-1\right)}{\frac{{\cos }^2\theta }{{\mathrm{a}}^2}+\frac{{\sin }^2\theta }{{\mathrm{b}}^2}}=1(-1)$

$\left(\frac{1}{a^2}+\frac{1}{b^2}\left(\frac{b^4{x}^2}{a^4{y}^2}\right)\right)=\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)\left(1+\frac{b^4{x}^2}{a^4{y}^2}\right)$ is the required locus

c) Equation of AB is $\frac{\mathrm{xh}}{{\mathrm{a}}^2}+\frac{\mathrm{yk}}{{\mathrm{b}}^2}=\frac{{\mathrm{h}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{k}}^2}{{\mathrm{b}}^2}$ Also AB is chord of contact of $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ w.r.t ellipse ⇒ Equation of AB is $\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}+\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}=1\dots$. (2)

∴ (1) and (2) are identical $\Rightarrow \frac{{\mathrm{x}}_1}{\mathrm{h}}=\frac{{\mathrm{y}}_1}{\mathrm{k}}=\frac{1}{\frac{{\mathrm{h}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{k}}^2}{{\mathrm{b}}^2}}$ $\left(x_1,y_1\right)$ lies on director circle $x^2+y^2=a^2+b^2$ of ellipse $\Rightarrow \frac{h^2+k^2}{\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}\right)}=a^2+b^2$ $\Rightarrow x^2+y^2=\left(a^2+b^2\right){\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}^2$ is the required locus.

d) Put $(2h-a,2k)$ to equation of ellipse $\Rightarrow \frac{(2h-a{)}^2}{a^2}+\frac{(2k{)}^2}{b^2}=1$

$\frac{{\left(x-\frac{a}{2}\right)}^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{4}$ is the required locus.