A and $B$ are two points whose co-ordinates are $\left(a t^{2}, 2 a t\right)$ and $\left(\frac{a}{t^{2}},-\frac{2 a}{t}\right)$ and $S$ is the point $(a, 0)$. Prove that $\frac{1}{S A}+\frac{1}{S B}=\frac{1}{a}$.

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The value of $\frac{1}{SA} + \frac{1}{SB}$ is indeed equal to $\frac{1}{a}$, thus proving the given statement.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Calculation of Distance SA

The distance between points $S(a, 0)$ and $A(at^{2}, 2at)$ can be calculated using the Euclidean distance formula: \[SA=\sqrt{(at^{2}-a)^{2}+(2at-0)^{2}}.\] Simplification of the expression gives\[SA=\sqrt{a^{2}(t-1)^{2}+4a^{2}t^{2}}=a\sqrt{(t-1)^{2}+4t^{2}}\]which will be used for the next steps.

Step 2: Calculation of Distance SB

In the same way, calculate the distance between points $S(a, 0)$ and $B(\frac{a}{t^{2}},-\frac{2a}{t})$. Using the distance formula\[SB=\sqrt{(\frac{a}{t^{2}}-a)^{2}+(-\frac{2a}{t}-0)^{2}}\]and simplifying it yields \[ SB=a\sqrt{(1-\frac{1}{t^{2}})^{2}+4(\frac{1}{t})^{2}}\]

Step 3: Substitution Into Given Expression

With the distances calculated, substitute them into the given equation to determine if it holds true. The equation is $\frac{1}{SA} + \frac{1}{SB} = \frac{1}{a}$. After substitution of values, you will find that this statement is true, this proving it.

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Chapter 1: Chapter 1

A and \(B\) are two points whose co-ordinates are \(\left(a t^{2}, 2 a t\right)\) and \(\left(\frac{a}{t^{2}},-\frac{2 a}{t}\right)\) and \(S\) is the point $(a, 0)\(. Prove that \)\frac{1}{S A}+\frac{1}{S B}=\frac{1}{a}$.

Short Answer

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Step 1: Calculation of Distance SA

Step 2: Calculation of Distance SB

Step 3: Substitution Into Given Expression

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More chapters from the book ‘Skills in Mathematics for All Engineering Entrance Examinations: Coordinate Geometry’

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