Question

Please explain how u got answer​

Answer 1

Answer:

$\frac{1}{p^2} =\frac{1}{a^2} +\frac{1}{b^2}$

Step-by-step explanation:

Converting the Intercept form to the General form

$\frac{x}{a} +\frac{y}{b} =1$

⇒ $\frac{bx+ay}{ab} =1$

⇒ $bx+ay=ab$

⇒ $bx+ay-ab=0$   Hence Converted

Perpendicular distance of a point  $P(x_1,y_1)$  to the line $Ax+By+C=0$ is:

                                                   $\frac{\lvert Ax_1+By_1+C \rvert}{\sqrt{A^2+B^2}}$

Here $(x_1,y_1)$  is  $(0,0)$ and $Ax+By+C=0$  is  $bx+ay-ab=0$

Substituting the values:

$\frac{\lvert b(0)+a(0)-ab \rvert}{\sqrt{b^2+a^2}}$

= $\frac{ab}{\sqrt{a^2+b^2}}$

Now perependicular distance  $p=\frac{ab}{\sqrt{a^2+b^2}}$

⇒ $p^2=\frac{a^2b^2}{a^2+b^2}$

⇒ $\frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2}$

⇒ $\frac{1}{p^2}=\frac{a^2}{a^2b^2}+\frac{b^2}{a^2b^2}$

⇒ $\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2}$

∴ Ans = $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 2

Step-by-step explanation:

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