Product of slopes of common tangents to the&n | vedclass

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Question

$y=m x+\frac{2}{m} ; y=m x \pm \sqrt{32 m^{2}+8}$

$	herefore \frac{2}{\mathrm{m}}=\pm \sqrt{32 \mathrm{m}^{2}+8}$

$\Rightarrow \frac{4}{m^{2}}=

Vedclass · Accepted Answer

$-\frac{1}{4}$

Vedclass · Answer

$y=m x+\frac{2}{m} ; y=m x \pm \sqrt{32 m^{2}+8}$

$	herefore \frac{2}{\mathrm{m}}=\pm \sqrt{32 \mathrm{m}^{2}+8}$

$\Rightarrow \frac{4}{m^{2}}=32 m^{2}+8$

$ \Rightarrow 1=8 m^{4}+2 m^{2}$

$\Rightarrow \mathrm{m}^{2}=-\frac{1}{2} ; \mathrm{m}^{2}=\frac{1}{4} $

$\Rightarrow \mathrm{m}=\frac{1}{2} \mathrm{or}-\frac{1}{2}$

$	herefore$ Product $=-\frac{1}{4}$

Product of slopes of common tangents to the ellipse $\frac{x^2}{32} + \frac{y^2}{8} = 1$ and parabola $y^2 = 8x$ is -

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