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Problem 13

# 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}$.

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

## 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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