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(B) The given equation of the hyperbola is $5 x^2-7 y^2-35=0$ ... $(i)$By differentiating $(i)$ with respect to $x$,we get $10 x-14 y \cdot y^{\prime}

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$36$

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(B) The given equation of the hyperbola is $5 x^2-7 y^2-35=0$ ... $(i)$By differentiating $(i)$ with respect to $x$,we get $10 x-14 y \cdot y^{\prime}=0$,which implies $y^{\prime}=\frac{5 x}{7 y}$.At the point $P(h, k)$,the slope of the tangent is $m = \frac{5 h}{7 k}$.Since the tangent is parallel to the line $\sqrt{2} x-y+\lambda=0$,its slope is $\sqrt{2}$.Thus,$\frac{5 h}{7 k} = \sqrt{2} \Rightarrow h = \frac{7 \sqrt{2} k}{5}$.Since $P(h, k)$ lies on the hyperbola,$5 h^2-7 k^2-35=0$.Substituting $h^2 = \frac{49 	imes 2 k^2}{25} = \frac{98 k^2}{25}$ into the hyperbola equation:$5 \left(\frac{98 k^2}{25}ight) - 7 k^2 = 35$ $\Rightarrow \frac{98 k^2}{5} - 7 k^2 = 35$ $\Rightarrow \frac{98 k^2 - 35 k^2}{5} = 35$ $\Rightarrow 63 k^2 = 175$ $\Rightarrow k^2 = \frac{175}{63} = \frac{25}{9}$.Since $P$ lies in the third quadrant,$k$ must be negative,so $k = -\frac{5}{3}$.Then $h^2 = \frac{98}{25} 	imes \frac{25}{9} = \frac{98}{9}$.Finally,$3 h^2 - 2 k = 3 \left(\frac{98}{9}ight) - 2 \left(-\frac{5}{3}ight) = \frac{98}{3} + \frac{10}{3} = \frac{108}{3} = 36$.

Let $P(h, k)$ be the point of contact of the tangent to the hyperbola $5 x^2-7 y^2-35=0$ which is parallel to the line $\sqrt{2} x-y+\lambda=0$. If $P$ lies in the third quadrant,then $3 h^2-2 k=$

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