Correct option is (D) \(\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\)
Let CD be the perpendicular from C to AB.
In right \(\triangle ACB,\)
\(\angle A+\angle B+\angle C=180^\circ\) (Sum of angles in a triangle)
\(\Rightarrow\) \(\angle A+\angle B=180^\circ-\angle C\)
\(\Rightarrow\) \(\angle A+\angle B=180^\circ-90^\circ\) \((\because\angle C=90^\circ)\)
\(\Rightarrow\) \(\angle A+\angle B=90^\circ\) ________________(1)
Let \(\angle ACD=\angle 1\;\&\;\angle BCD=\angle 2\)
Now in right \(\triangle ACD,\) we have
\(\angle A+\angle D+\angle 1=180^\circ\)
\(\Rightarrow\angle A+\angle 1=180^\circ-90^\circ=90^\circ\) \((\because\angle D=90^\circ\;(as\,CD\bot AB))\)
\(\Rightarrow\angle A+\angle 1=90^\circ\) ________________(2)
From (1) & (2), we get
\(\angle B=\angle 1\), similarly, \(\angle A=\angle 2\) ________________(3)
Now in triangles \(\triangle ADC\;and\;\triangle CDB,\)
\(\angle 1=\angle B\) (From (3))
\(\angle A=\angle 2\) (From (3))
\(\angle ADC=\angle CDB=90^\circ\) \((\because CD\bot AB)\)
\(\therefore\) \(\triangle ADC\sim\triangle CDB\)
\(\therefore\frac{AD}{CD}=\frac{CD}{BD}=\frac{AC}{BC}\) (By properties of similar triangles)
\(\Rightarrow\) \(\frac{AD}{CD}=\frac{CD}{BD}\)
\(\Rightarrow CD^2=AD.BD\)
\(\Rightarrow P^2=AD.(AB-AD)\) \((\because CD=P\;and\;BD=AB-AD)\)
\(\Rightarrow P^2=AB.AD-AD^2\) ________________(4)
Now in \(\triangle ADC,\) we have
\(AC^2=AD^2+CD^2\)
\(\Rightarrow AD^2=AC^2-CD^2\)
\(\Rightarrow AD^2=b^2-P^2\) ________________(5) \((\because AC=b\;\&\;CD=p)\)
From (4) & (5), we obtain
\(P^2=C.\sqrt{b^2-p^2}-(b^2-p^2)\) \((\because AB=c)\)
\(\Rightarrow p^2+b^2-p^2=c\sqrt{b^2-p^2}\)
\(\Rightarrow\frac{b^2}{c}=\sqrt{b^2-p^2}\)
\(\Rightarrow\frac{b^4}{c^2}=b^2-p^2\) (By squaring both sides)
\(\Rightarrow p^2=b^2-\frac{b^4}{c^2}\)
\(=b^2-\frac{b^4}{a^2+b^2}\) \((\because\,In\,right\,\triangle ACB,C^2=a^2+b^2)\)
\(=\frac{a^2b^2+b^4-b^4}{a^2+b^2}\) \(=\frac{a^2b^2}{a^2+b^2}\)
\(\Rightarrow\frac1{p^2}\) \(=\frac{a^2+b^2}{a^2b^2}\)
\(\Rightarrow\frac1{p^2}\) \(=\frac1{b^2}+\frac1{a^2}\)
\(\Rightarrow\frac1{p^2}\) \(=\frac1{a^2}+\frac1{b^2}\)
Alternative method :-
\(\because\) Area of triangle \(=\frac12\times\) Base \(\times\) corresponding altitude
If base is AC then its corresponding altitude is BC
\(\therefore\) Area of \(\triangle ABC\) \(=\frac12\times AC\times BC\)
\(=\frac12\times b\times a\) \(=\frac{ab}2\) ________________(1) \((\because AC=b\;\&\;BC=a)\)
If base is AB then its corresponding altitude is CD.
\(\therefore\) Area of \(\triangle ABC\) \(=\frac12\times AB\times CD\)
\(=\frac12\times c\times p\) \(=\frac{cp}2\) ________________(2) \((\because AB=c\;\&\;CD=p)\)
Both (1) & (2) represent the area of \(\triangle ABC\)
So, both must be equal.
\(\therefore\frac{ab}2=\frac{cp}2\)
\(\Rightarrow cp=ab\)
\(\Rightarrow c^2p^2=a^2b^2\) (By squaring both sides)
\(\Rightarrow\frac1{p^2}=\frac{c^2}{a^2b^2}\)
\(\Rightarrow\frac1{p^2}=\frac{a^2+b^2}{a^2b^2}\) (By applying Pythagoras theorem in \(\triangle ACB,c^2=a^2+b^2)\)
\(\Rightarrow\) \(\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\)
