If the straight line 8x + 3sqrt{2}y = 36 touc | vedclass

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(D) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{8} = 1$.Let the point of contact be $(a, b)$. The equation of the tangent at $(a, b)$

Vedclass · Accepted Answer

$\frac{16}{3}$

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(D) The equation of the ellipse is $\frac{x^2}{18} + \frac{y^2}{8} = 1$.Let the point of contact be $(a, b)$. The equation of the tangent at $(a, b)$ to the ellipse $\frac{x^2}{18} + \frac{y^2}{8} = 1$ is $\frac{ax}{18} + \frac{by}{8} = 1$.Given the tangent line is $8x + 3\sqrt{2}y = 36$,which can be written as $\frac{8x}{36} + \frac{3\sqrt{2}y}{36} = 1$,or $\frac{2x}{9} + \frac{\sqrt{2}y}{12} = 1$.Comparing the two equations of the tangent:$\frac{a}{18} = \frac{2}{9} \implies a = \frac{18 	imes 2}{9} = 4$.$\frac{b}{8} = \frac{\sqrt{2}}{12} \implies b = \frac{8\sqrt{2}}{12} = \frac{2\sqrt{2}}{3}$.Now,calculate $a + \sqrt{2}b = 4 + \sqrt{2} 	imes \frac{2\sqrt{2}}{3} = 4 + \frac{4}{3} = \frac{12 + 4}{3} = \frac{16}{3}$.

If the straight line $8x + 3\sqrt{2}y = 36$ touches the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 2$ at $(a, b)$,then $a + \sqrt{2}b =$

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