Hyperbola in conic sections

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$A normal to the hyperbola \frac{x^{2}}{4} - \frac{y^{2}}{1} = 1 has equal intercepts on the positive x - and y -axes. If this normal touches the$
$ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, then a^{2} + b^{2} is equal to$

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Solution

$The equation of the normal to the hyperbola \frac{x^{2}}{4} - \frac{y^{2}}{1} = 1$
$\begin{array}{l} at (2 \sec θ, \tan θ) is 2 x \cos θ + y \cot θ = 5 . \\ The slope of the normal is \end{array}$
$\begin{array}{l} - 2 \sin θ = - 1 \\ or \sin θ = \frac{1}{2} or θ = \frac{π}{6} \\ Y -intercept of the normal = \frac{5}{\cot θ} = \frac{5}{\sqrt{3}} \end{array}$
$\begin{array}{l} Y -intercept of the normal = \frac{5}{\cot θ} = \frac{5}{\sqrt{3}} \\ As it touches the ellipse \end{array}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$we have a^{2} + b^{2} = \frac{25}{3} (Using a^{2} m^{2} + b^{2} = c^{2})$

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Hyperbola in conic sections

Remember concepts with our Masterclasses.

A normal to the hyperbola x24−y21=1 has equal intercepts on the positive x - and y -axes. If this normal touches the ellipse x2a2+y2b2=1, then a2+b2 is equal to

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$A normal to the hyperbola \frac{x^{2}}{4} - \frac{y^{2}}{1} = 1 has equal intercepts on the positive x - and y -axes. If this normal touches the$
$ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, then a^{2} + b^{2} is equal to$