Hyperbola in conic sections

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$A normal to the hyperbola \frac{x^{2}}{6} - \frac{y^{2}}{2} = 1 has equal intercepts on positive x and y axes. If this normal touches$
$the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, then the value of (a^{2} + b^{2}) is$

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Solution

$The equation of normal to hyperbola$
$\begin{array}{l} \frac{x^{2}}{6} - \frac{y^{2}}{2} = 1 at (\sqrt{6} \sec θ, \sqrt{2} \tan θ) is \\ \frac{\sqrt{6} x}{\sec θ} + \frac{\sqrt{2} y}{\tan θ} = 8 \end{array}$
$\begin{array}{l} Slope = - 1 \\ \Rightarrow \frac{- \sqrt{6}}{\sec θ} \times \frac{\tan θ}{\sqrt{2}} = - 1 \\ \Rightarrow \sin θ = \frac{1}{\sqrt{3}} \end{array}$
$\begin{array}{l} Thus, equation of normal is x + y = 4 \\ So, y = - x + 4 is tangent to \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ Hence, (a^{2} + b^{2}) = 16 \end{array}$

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Hyperbola in conic sections

Remember concepts with our Masterclasses.

A normal to the hyperbola x26−y22=1 has equal intercepts on positive x and y axes. If this normal touches the ellipse x2a2+y2b2=1, then the value of a2+b2 is

The equation of normal to hyperbolax26−y22=1 at (6sec⁡θ,2tan⁡θ) is 6xsec⁡θ+2ytan⁡θ=8 Slope =−1⇒ −6sec⁡θ×tan⁡θ2=−1⇒ sin⁡θ=13 Thus, equation of normal is x+y=4 So, y=−x+4 is tangent to x2a2+y2b2=1 Hence, a2+b2=16

Ready to Test Your Skills?

free study material

$A normal to the hyperbola \frac{x^{2}}{6} - \frac{y^{2}}{2} = 1 has equal intercepts on positive x and y axes. If this normal touches$
$the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, then the value of (a^{2} + b^{2}) is$