Hyperbola in conic sections

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Question

$Number of integral values of b for which tangent parallel to line y = x + 1 can be drawn to hyperbola \frac{x^{2}}{5} - \frac{y^{2}}{b^{2}} = 1 is$

Moderate

Solution

$Equation of tangent to hyperbola \frac{x^{2}}{5} - \frac{y^{2}}{b^{2}} = 1 having slope m is$
$\begin{aligned} y = m x \pm \sqrt{a^{2} m^{2} - b^{2}} \\ or y = x \pm \sqrt{5 - b^{2}} \end{aligned}$
$Comparing it with y = x + 1, we get$
$\begin{aligned} b^{2} = 1, 4 \\ ∴ b = \pm 1, \pm 2 \end{aligned}$
$Thus, four integral values of b are possible.$

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Similar Questions

Statement-1: The locus of the point of intersection of the tangents that are at right angles to the hyperbola $\frac{x^{2}}{36} - \frac{y^{2}}{16} = 1 is the circle x^{2} + y^{2} = 52$

Statement-2: Perpendicular tangents to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ interest on the director circle $x^{2} + y^{2} = a^{2} - b^{2} (a^{2} > b^{2})$ of the hyperbola.