If two tangents are drawn to the hyperbola fr | vedclass

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(A) Let $(h, k)$ be the point of intersection of the two tangents.The equation of the pair of tangents from $(h, k)$ to the hyperbola is given by $SS_

Vedclass · Accepted Answer

$y^2 + b^2 = c^2(x^2 - a^2)$

Vedclass · Answer

(A) Let $(h, k)$ be the point of intersection of the two tangents.The equation of the pair of tangents from $(h, k)$ to the hyperbola is given by $SS_1 = T^2$,where $S = \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1$,$S_1 = \frac{h^2}{a^2} - \frac{k^2}{b^2} - 1$,and $T = \frac{hx}{a^2} - \frac{ky}{b^2} - 1$.Expanding $SS_1 = T^2$ gives a quadratic equation in $x$ and $y$ of the form $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$.The product of the slopes $m_1m_2$ of the lines represented by this quadratic equation is given by $\frac{A}{B}$.Calculating the coefficients of $x^2$ and $y^2$ from the expansion:$A = \frac{1}{a^2}(\frac{h^2}{a^2} - \frac{k^2}{b^2} - 1) - \frac{h^2}{a^4} = -\frac{k^2}{a^2b^2} - \frac{1}{a^2}$$B = -\frac{1}{b^2}(\frac{h^2}{a^2} - \frac{k^2}{b^2} - 1) - \frac{k^2}{b^4} = -\frac{h^2}{a^2b^2} + \frac{1}{b^2}$Given $m_1m_2 = c^2$,we have $\frac{-\frac{k^2}{a^2b^2} - \frac{1}{a^2}}{-\frac{h^2}{a^2b^2} + \frac{1}{b^2}} = c^2$.Simplifying this,we get $\frac{k^2 + b^2}{h^2 - a^2} = c^2$,which implies $y^2 + b^2 = c^2(x^2 - a^2)$.

If two tangents are drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that the product of their gradients is $c^2$,then they intersect on the curve:

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