Question

Coordinates of the focus of the parabola $\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$ is

• A. $(\frac{ab}{a+b},\frac{ab}{a+b})$
• B. $(\frac{a{b}^2}{a^2+b^2},\frac{a^{2}b}{a^2+b^2})$
• C. $(\frac{a^{2}b}{a+b},\frac{a{b}^2}{a+b})$
• D. $(a,b)$

Answer 1

The correct option is B $(\frac{a{b}^2}{a^2+b^2},\frac{a^{2}b}{a^2+b^2})$

$\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1$
For this parabola x axis is a tangent at P(a, 0)
Y-axis a tangent Q(0,b)
$\therefore$ O(0,0) is point if inter section perpendicular tangents $\therefore$ Directrix passing through this point
Clearly $\angle OSP={90}^{\circ }$
Hence circle on OP as diameter passing though S
i.e., $x^2+y^2-ax=0$ passing through S.
similarly, $\angle OSQ={90}^{\circ }\therefore x^2+y^2-bx=0$ passing through S.
Point of intersection of above circles is focus.
$x^2+y^2-ax=0x^2+y^2-bx=0\underset{–}{}ax-by=0$
$y=\frac{ax}{b}\Rightarrow x^2+\frac{a^2{x}^2}{b^2}=ax$
$x\left(\frac{b^2+a^2}{b^2}\right)=ax=\frac{a{b}^2}{a^2+b^2}y=\frac{a^{2}b}{a^2+b^2}$
Focus $S=(\frac{a{b}^2}{a^2+b^2},\frac{a^{2}b}{a^2+b^2})$.