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Question

The slope of the line touching both the parabolas $$y^{2}=4x$$ and $$ x^{2}=-32y$$ is


A
12
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B
32
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C
18
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D
23
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Solution

The correct option is A $$ \displaystyle \frac{1}{2}$$
Equation of tangent at $$ A(t^{2},2t)$$

$$ yt=x+t^{2}$$ is tangent to $$ x^{2}+32y=0$$ at $$ B$$

$$ \Rightarrow x^{2}+32(\displaystyle \frac{x}{t}+t)=0$$

$$ \Rightarrow x^{2}+\displaystyle \frac{32}{t}x+32t=0$$

$$ \Rightarrow ({\displaystyle \frac{32}{t})^{2}}-4(32t)=0$$

$$ \Rightarrow 32(\displaystyle \frac{32}{t^{2}}-4t)=0$$

$$ \Rightarrow t^{3}=8\Rightarrow t=2$$

$$ \Rightarrow$$ Slope of tangent is $$ \displaystyle \frac{1}{t}$$=$$\displaystyle \frac{1}{2}$$

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Mathematics

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