Given ellipse $x^{2} + 4 y^{2} = 16$ and parabola $y^{2} - 4 x - 4 = 0.$ The quadratic equation whose roots are squares of the slopes of the common tangent to parabola and ellipse, is

$3 x^{2} - 1 = 0$

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$5 x^{2} - 1 = 0$

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$15 x^{2} + 2 x - 1 = 0$

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$2 x^{2} - 1 = 0$

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Solution

The correct option is C
$15 x^{2} + 2 x - 1 = 0$

Ellipse: $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1; p a r a b o l a : y^{2} = 4 (x + 1) P u t x + 1 = X, Y = y; y^{2} = 4 X, s o t a n g e n t, i s Y = m X + \frac{1}{m} = m (x + 1) + \frac{1}{m} y = m x + (m + \frac{1}{m}) . . . . . . . . (1) I f (1) i s a l s o a t a n g e n t o n \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1 t h e n c^{2} = a^{2} m^{2} + b^{2} \Rightarrow {(m + \frac{1}{m})}^{2} = 16 m^{2} + 4 \Rightarrow m^{2} + \frac{1}{m^{2}} + 2 = 16 m^{2} + 4 \Rightarrow 15 m^{2} - \frac{1}{m^{2}} + 2 = 0 L e t m^{2} = t > 0 \Rightarrow 15 t^{2} + 2 t - 1 = 0$

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Similar questions

x^{2} + 4 y^{2} = 16

y^{2} - 4 x - 4 = 0.

x^{2} + 4 y^{2} = 16

y^{2} - 4 x - 4 = 0.

\frac{x^{2}}{36} + \frac{y^{2}}{9} = 1

y^{2} = 4 x .

y^{2} = 16 \sqrt{3} x

2 x^{2} + y^{2} = 4

y = 2 x + 2 \sqrt{3}

y = m x + \frac{4 \sqrt{3}}{m}, (m \neq 0)

y^{2} = 16 \sqrt{3} x

2 x^{2} + y^{2} = 4

m

m^{4} + 2 m^{2} = 24

The equation whose roots are the squares of the roots of the equation
$2 x^{2} + 3 x + 1 = 0$ is

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