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  • If p is the length of perpendicular from the origin on the line x/a+y/b=1 and a2, p2, b2 are in A.P, then show that a4 + b4=0.

If p is the length of the perpendicular from the origin on the line \frac{x}{a}+\frac{y}{b}=1 and a2, p2, b2 are in A.P, then show that a4 + b4 = 0.

Answers (1)

Equation of line in intercept form=\frac{x}{a}+\frac{y}{b}=1

  Since, p is the length of perpendicular drawn from the origin to the given line, p= \left | \frac{\frac{0}{a}+\frac{0}{b}-1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}} \right |  

Squaring both sides, we have p= \left | \frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}} \right |

  \frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}........(i)

Since, a2,b2  and p2  are in AP

  2p2=a2+b2  

p^{2}=\frac{\left ( a^{2}+b^{2} \right )}{2}

\frac{1}{p^{2}}=\frac{2}{a^{2}+b^{2}}..........(ii)

 From equation (i) and (ii)  we get   \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{2}{a^{2}+b^{2}}

\frac{b^{2}+a^{2}}{a^{2}b^{2}}=\frac{2}{a^{2}+b^{2}}

\left ( a^{2}+b^{2} \right )\left ( a^{2}+b^{2} \right )=2\left ( a^{2}b^{2} \right )

a^{4}+b^{4}=0

Hence, proved. 

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