Question

A variable plane cuts off intercepts from the co-ordinate axes which are equal to the roots of the equation $x^3+5x=p-q{x}^2$ (p,q are real numbers).The locus of the foot of the perpendicular from the origin to the plane is

• A. $(x^2+y^2+z^2{)}^2(xy+yz+zx)=5$
• B. $(x^2+y^2+z^2{)}^4(1/xy+1/yz+1/zx)=5$
• C. $(x^2+y^2+z^2{)}^2(1/xy+1/yz+1/zx)=5$
• D. $(x^2+y^2+z^2{)}^4(xy+yz+zx)=5$

Answer 1

Answer: (C)

$(x^2+y^2+z^2{)}^2(1/xy+1/yz+1/zx)=5$

Let a, b, c be the intercepts cut off by the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ or $lx+my+nz=p$ on the coordinate axis

We know $a=\frac{p}{l},b=\frac{p}{m},c=\frac{p}{n}$ where p is length of perpendicular from origin to the plane and l, m, n are the direction cosines of the normals.

Foot of perpendicular from origin to the plane is $(\alpha ,\beta ,\gamma )=(pl,pm,pn)$

Now $ab+bc+ca=5$

$\Rightarrow p^4(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })=5\Rightarrow ({\alpha }^2+{\beta }^2+{\gamma }^2{)}^2(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })=5$

$\Rightarrow$ Locus is $(x^2+y^2+z^2{)}^2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})=5$