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  • For , the circle <img alt="\mathrm{x^2+y^2=1}" src="https://entrancecorner.oncodecogs.com/gif

For \mathrm{a>0}, the circle \mathrm{x^2+y^2=1} and the parabola \mathrm{y=a x^2-b} intersect in how many distinct points provided \mathrm{a>b>1.}

Option: 1

0


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

The given circle is \mathrm{x^2+y^2=1}. Its centre is at (0,0) and its radius is 1 . The given parabola is \mathrm{y+b=a x^2} whose vertex is at \mathrm{(0,-b)}. The circle is symmetrical about the x-axis and the y-axis. The parabola is symmetrical about the y-axis. Since the four points of intersection are distinct, the vertex of the parabola must be outside the circle.

\mathrm{\Rightarrow b>1}. Also the points of intersection are given by \mathrm{\frac{y+b}{a}+y^2=1}

\mathrm{ \Rightarrow a y^2+y+b-a=0}

One root of this equation is positive and the other is negative.

Hence \mathrm{\frac{\mathrm{b}-\mathrm{a}}{\mathrm{a}}<0 \Rightarrow \mathrm{b}<\mathrm{a}\, \, as \, \, \mathrm{a}} is positive.

Hence \mathrm{a>b>1.}

Posted by

Divya Prakash Singh

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