Question

If A & B are point on the parabola $y^2=4ax$ with vertex O such that OA perpendicular to OB & having lengths $r_1$ & $r_2$ respectively, then the value of $\frac{r_1^{4/3}{r}_2^{4/3}}{r_1^{2/3}+r_2^{2/3}}$

• A. $16a^2$
• B. $a^2$
• C. 4a
• D. None of these

Answer 1

The correct option is A $16a^2$

Let coordinates of point A are $A(r_{1}cos\theta ,r_{1}sin\theta )$

Thus, coordinates of point B will be $B(r_{2}cos(90-\theta ),r_{2}sin(90-\theta ))$
$=B(r_{2}sin\theta ,r_{2}cos\theta )$

Now, equation of parabola is $y^2=4ax$

Point A lies on the parabola. Thus it must satisfy the equation of parabola.

$\therefore {\left(r_{1}sin\theta \right)}^2=4a\left(r_{1}cos\theta \right)$

$\therefore r_1^2{sin}^2\theta =4a{r}_{1}cos\theta$

$\therefore r_1{sin}^2\theta =4acos\theta$ (1)

Squaring both sides, we get,

$\therefore {r_1}^2{sin}^4\theta =16a^2{cos}^2\theta$ (2)

Point B also lies on the parabola. Thus it must satisfy the equation of parabola.

$\therefore {\left(r_{2}cos\theta \right)}^2=4a\left(r_{2}sin\theta \right)$

$\therefore {r_2}^2{cos}^2\theta =4a{r}_{2}sin\theta$

$\therefore r_2{cos}^2\theta =4asin\theta$ (3)

$\therefore {cos}^2\theta =\frac{4asin\theta }{r_2}$

Putting this value in equation (1), we get,

${r_1}^2{sin}^4\theta =16a^2\times \frac{4asin\theta }{r_2}$

${r_1}^2{sin}^3\theta =\frac{64a^3}{r_2}$

${sin}^3\theta =\frac{64a^3}{{r_1}^2{r}_2}$

$\therefore {\left({sin}^3\theta \right)}^{2/3}={\left(\frac{64a^3}{{r_1}^2{r}_2}\right)}^{2/3}$

$\therefore {sin}^2\theta =\frac{16a^2}{{r_1}^{4/3}{r_2}^{2/3}}$ (4)

Taking ratio of equations (1) and (3), we get,

$\frac{r_1{sin}^2\theta }{r_2{cos}^2\theta }=\frac{4acos\theta }{4asin\theta }$

$\therefore r_1{sin}^2\theta \times 4asin\theta =r_2{cos}^2\theta \times 4acos\theta$

$\therefore r_1{sin}^3\theta =r_2{cos}^3\theta$

$\therefore {cos}^3\theta =\frac{r_1}{r_2}{sin}^3\theta$

$\therefore {cos}^3\theta =\frac{r_1}{r_2}\frac{64a^3}{{r_1}^2{r}_2}$

$\therefore {cos}^3\theta =\frac{64a^3}{r_1{r_2}^2}$

$\therefore {\left({cos}^3\theta \right)}^{2/3}={\left(\frac{64a^3}{r_1{r_2}^2}\right)}^{2/3}$

$\therefore {cos}^2\theta =\frac{16a^2}{{r_1}^{2/3}{r_2}^{4/3}}$ (5)

Adding equations (4) and (5), we get,

${sin}^2\theta +{cos}^2\theta =\frac{16a^2}{{r_1}^{4/3}{r_2}^{2/3}}+\frac{16a^2}{{r_1}^{2/3}{r_2}^{4/3}}$

$\therefore 1=16a^2[\frac{1}{{r_1}^{4/3}{r_2}^{2/3}}+\frac{1}{{r_1}^{2/3}{r_2}^{4/3}}]$

$\therefore 1=\frac{16a^2}{{r_1}^{2/3}{r_2}^{2/3}}[\frac{1}{{r_1}^{2/3}}+\frac{1}{{r_2}^{2/3}}]$

$\therefore 1=\frac{16a^2}{{r_1}^{2/3}{r_2}^{2/3}}\left[\frac{{r_1}^{2/3}+{r_2}^{2/3}}{{r_1}^{2/3}{r_2}^{2/3}}\right]$

$\therefore 1=\frac{16a^2}{{r_1}^{4/3}{r_2}^{4/3}}[{r_1}^{2/3}+{r_2}^{2/3}]$

$\therefore \frac{{r_1}^{4/3}{r_2}^{4/3}}{{r_1}^{2/3}+{r_2}^{2/3}}=16a^2$