JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q28

← JEE Main 2022 · 25 Jul · Shift 1

MathematicsInteger+4 / -1Question 28 of 90

Let the equation of two diameters of a circle $x^2+y^2-2x+2fy+1=0$ be $2px-y=1$ and $2x+py=4p$ . Then the slope $m\in(0,\infty)$ of the tangent to the hyperbola $3x^2-y^2=3$ passing through the centre of the circle is equal to ______.

Your answer

Answer and worked solution

Official answer: 2

Centre $(1,-f)$ lies on both diameters: $p=2$ , $f=-3$ , centre $(1,3)$ (the other root gives a point circle).

Tangent $y-3=m(x-1)$ to $x^2-\frac{y^2}3=1$ : $(3-m)^2=m^2-3$ , so $m=2$ .

Answer: 2

Topics:Tangents and normals to conics Equation of a circle

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JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 28