Y-4x2 and x2-a2-y2-16-1 intersect- iff -

The correct option is A |a|≤1/√(2) x^2=y4 y4x^2 y^216=1 a^2y^2 4y+16a^2=0 d gt;0 4^2 16× 4a^4 gt;0 a^4 lt;14 |a| ...

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Algebra of Derivatives

y=4x2 and x...

Question

$y = 4 x^{2}$ and $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{16} = 1$ intersect, iff :

| a | \leq \frac{1}{\sqrt{2}}

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a > \frac{1}{\sqrt{2}}

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a > - \frac{1}{\sqrt{2}}

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a > \sqrt{2}

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Solution

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List- I	List- II
A. The locus of the point of intersection of the perpendicular tangents of $x^{2} + y^{2} = a^{2}$	1) $x^{2} + y^{2} = 5$ of
B. The locus of the mid-point of the chord of the circle $x^{2} + y^{2} = 8$ which is $\sqrt{2}$ units from origin is	2) $x^{2} + y^{2} = a^{2} {sec}^{2} α$
C. The locus of the point of intersection of the lines $x = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2 t}{1 + t^{2}}$	3) $x^{2} + y^{2} = 2 a^{2}$
D. The locus of the point whose chord of contact w.r.t $x^{2} + y^{2} = a^{2}$ making an angle '2 $α$ ' at the center is	4) $x^{2} + y^{2} = 2$ 5) $x^{2} + y^{2} = 1$

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