A) \[180{}^\circ \]
B) \[90{}^\circ \]
C) \[0{}^\circ \]
D) None of these
Correct Answer: B
Solution :
[b] We have, \[{{y}^{2}}=2ax\] Put \[x=\frac{a}{2};{{y}^{2}}=2a\left( \frac{a}{2} \right)\Rightarrow y=\pm a\] \[\therefore \] The points are \[\left( \frac{a}{2},a \right)\] and \[\left( \frac{a}{2},-a \right)\] Differentiating (i) with respect to x, we get \[2y\frac{dy}{dx}=2a\Rightarrow \frac{dy}{dx}=\frac{a}{y}\] At \[\left( \frac{a}{2},a \right);\frac{dy}{dx}=\frac{a}{y}=\frac{a}{a}=1={{m}_{1}}\,\,\,(say)\] At \[\left( \frac{a}{2},-a \right);\frac{dy}{dx}=\frac{a}{y}=\frac{a}{-a}=-1={{m}_{2}}\,\,\,\,\,\,(say)\] Since \[{{m}_{1}}{{m}_{2}}=-1\], the two tangents are at right angles.
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