Question

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

A
12
B
32
C
18
D
23

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 $$\displaystyle \frac{1}{t}$$=$$\displaystyle \frac{1}{2}$$Mathematics

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