If the variable line y = kx + 2h is tangent t | vedclass

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(D) The equation of the ellipse is $2x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{3} + \frac{y^2}{2} = 1$. Here,$a^2 = 3$ and $b^2 = 2$. The li

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$\sqrt{\frac{7}{3}}$

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(D) The equation of the ellipse is $2x^2 + 3y^2 = 6$,which can be written as $\frac{x^2}{3} + \frac{y^2}{2} = 1$. Here,$a^2 = 3$ and $b^2 = 2$. The line $y = kx + 2h$ is tangent to the ellipse. The condition for the line $y = mx + c$ to be tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$. Comparing $y = kx + 2h$ with $y = mx + c$,we have $m = k$ and $c = 2h$. Substituting these into the condition: $(2h)^2 = 3(k)^2 + 2$,which simplifies to $4h^2 = 3k^2 + 2$. Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $4x^2 - 3y^2 = 2$,or $\frac{x^2}{1/2} - \frac{y^2}{2/3} = 1$. This is a hyperbola with $a^2 = \frac{1}{2}$ and $b^2 = \frac{2}{3}$. The eccentricity $e$ of a hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$. $e = \sqrt{1 + \frac{2/3}{1/2}} = \sqrt{1 + \frac{4}{3}} = \sqrt{\frac{7}{3}}$.

If the variable line $y = kx + 2h$ is tangent to an ellipse $2x^2 + 3y^2 = 6$,then the locus of $P(h, k)$ is a conic $C$ whose eccentricity equals

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