If the product of the slopes of the tangents | vedclass

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(A) Let the point $P$ be $(h, k)$. The equation of a line passing through $(h, k)$ with slope $m$ is $y - k = m(x - h)$,or $y = mx + (k - mh)$.For thi

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$y^2+b^2=k^2\left(x^2-a^2\right)$

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(A) Let the point $P$ be $(h, k)$. The equation of a line passing through $(h, k)$ with slope $m$ is $y - k = m(x - h)$,or $y = mx + (k - mh)$.For this line to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition of tangency is $c^2 = a^2m^2 - b^2$,where $c = k - mh$.Substituting $c$,we get $(k - mh)^2 = a^2m^2 - b^2$.Expanding this,$k^2 - 2mhk + m^2h^2 = a^2m^2 - b^2$.Rearranging as a quadratic in $m$: $m^2(h^2 - a^2) - 2mhk + (k^2 + b^2) = 0$.Let the slopes of the two tangents be $m_1$ and $m_2$. The product of the roots is $m_1m_2 = \frac{k^2 + b^2}{h^2 - a^2}$.Given that the product of the slopes is $k^2$,we have $\frac{k^2 + b^2}{h^2 - a^2} = k^2$.Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $\frac{y^2 + b^2}{x^2 - a^2} = k^2$,which simplifies to $y^2 + b^2 = k^2(x^2 - a^2)$.

If the product of the slopes of the tangents drawn from an external point $P(h, k)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is a constant $k^2$,then the locus of $P$ is

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