Let a line L1 be tangent to the hyperbola
\(\frac{x²}{16} - \frac{y²}{4} = 1\)
and let L2 be the line passing through the origin and perpendicular to L1. If the locus of the point of intersection of L1 and L2 is
\(( x² + y²)² = αx² + βy²,\)
then α + β is equal to___.
Let a line L1 be tangent to the hyperbola
\(\frac{x²}{16} - \frac{y²}{4} = 1\)
and let L2 be the line passing through the origin and perpendicular to L1. If the locus of the point of intersection of L1 and L2 is
\(( x² + y²)² = αx² + βy²,\)
then α + β is equal to___.
Correct Answer: 12
Solution and Explanation
The correct answer is 12
Equation of L1 is
\(\frac{xsecθ}{4} - \frac{ytanθ}{2} = 1 ...... (i)\)
Equation of line L2 is
\(\frac{x tanθ}{2} + \frac{y secθ}{4} = 0 ....... (ii)\)
∵ Required point of intersection of L1 and L2 is (x1, y1) then
\(\frac{x_1secθ}{4} - \frac{y_1tanθ}{2} - 1 = 0 ...... (iii)\)
and
\(\frac{y_1secθ}{4} + \frac{x_1tanθ}{2} = 0 ....... (iv)\)
From equations (iii) and (iv), we get
\(\sec\theta = \frac{4x_1}{x_1^2 + y_1^2}\) and \(\tan\theta = \frac{-2y_1}{x_1^2 + y_1^2}\)
∴ Required locus of (x1, y1) is
\(( x² + y² )² = 16x² - 4y²\)
∴ α = 16, β = -4
Therefore, α + β = 12
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
